Find the Local Maxima and Minima y=-x^4-20x^3-148x^2-480x-576

$y = - x^{4} - 20 x^{3} - 148 x^{2} - 480 x - 576$

Step 1

Write $y = - x^{4} - 20 x^{3} - 148 x^{2} - 480 x - 576$ as a function.

$f (x) = - x^{4} - 20 x^{3} - 148 x^{2} - 480 x - 576$

Step 2

Find the first derivative of the function.

By the Sum Rule, the derivative of $- x^{4} - 20 x^{3} - 148 x^{2} - 480 x - 576$ with respect to $x$ is $\frac{d}{d x} [- x^{4}] + \frac{d}{d x} [- 20 x^{3}] + \frac{d}{d x} [- 148 x^{2}] + \frac{d}{d x} [- 480 x] + \frac{d}{d x} [- 576]$ .

$\frac{d}{d x} [- x^{4}] + \frac{d}{d x} [- 20 x^{3}] + \frac{d}{d x} [- 148 x^{2}] + \frac{d}{d x} [- 480 x] + \frac{d}{d x} [- 576]$

Evaluate $\frac{d}{d x} [- x^{4}]$ .

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{4}$ with respect to $x$ is $- \frac{d}{d x} [x^{4}]$ .

$- \frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 20 x^{3}] + \frac{d}{d x} [- 148 x^{2}] + \frac{d}{d x} [- 480 x] + \frac{d}{d x} [- 576]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$- (4 x^{3}) + \frac{d}{d x} [- 20 x^{3}] + \frac{d}{d x} [- 148 x^{2}] + \frac{d}{d x} [- 480 x] + \frac{d}{d x} [- 576]$

Multiply $4$ by $- 1$ .

$- 4 x^{3} + \frac{d}{d x} [- 20 x^{3}] + \frac{d}{d x} [- 148 x^{2}] + \frac{d}{d x} [- 480 x] + \frac{d}{d x} [- 576]$

Evaluate $\frac{d}{d x} [- 20 x^{3}]$ .

Since $- 20$ is constant with respect to $x$ , the derivative of $- 20 x^{3}$ with respect to $x$ is $- 20 \frac{d}{d x} [x^{3}]$ .

$- 4 x^{3} - 20 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 148 x^{2}] + \frac{d}{d x} [- 480 x] + \frac{d}{d x} [- 576]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$- 4 x^{3} - 20 (3 x^{2}) + \frac{d}{d x} [- 148 x^{2}] + \frac{d}{d x} [- 480 x] + \frac{d}{d x} [- 576]$

Multiply $3$ by $- 20$ .

$- 4 x^{3} - 60 x^{2} + \frac{d}{d x} [- 148 x^{2}] + \frac{d}{d x} [- 480 x] + \frac{d}{d x} [- 576]$

Evaluate $\frac{d}{d x} [- 148 x^{2}]$ .

Since $- 148$ is constant with respect to $x$ , the derivative of $- 148 x^{2}$ with respect to $x$ is $- 148 \frac{d}{d x} [x^{2}]$ .

$- 4 x^{3} - 60 x^{2} - 148 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 480 x] + \frac{d}{d x} [- 576]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$- 4 x^{3} - 60 x^{2} - 148 (2 x) + \frac{d}{d x} [- 480 x] + \frac{d}{d x} [- 576]$

Multiply $2$ by $- 148$ .

$- 4 x^{3} - 60 x^{2} - 296 x + \frac{d}{d x} [- 480 x] + \frac{d}{d x} [- 576]$

Evaluate $\frac{d}{d x} [- 480 x]$ .

Since $- 480$ is constant with respect to $x$ , the derivative of $- 480 x$ with respect to $x$ is $- 480 \frac{d}{d x} [x]$ .

$- 4 x^{3} - 60 x^{2} - 296 x - 480 \frac{d}{d x} [x] + \frac{d}{d x} [- 576]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$- 4 x^{3} - 60 x^{2} - 296 x - 480 \cdot 1 + \frac{d}{d x} [- 576]$

Multiply $- 480$ by $1$ .

$- 4 x^{3} - 60 x^{2} - 296 x - 480 + \frac{d}{d x} [- 576]$

Differentiate using the Constant Rule.

Since $- 576$ is constant with respect to $x$ , the derivative of $- 576$ with respect to $x$ is $0$ .

$- 4 x^{3} - 60 x^{2} - 296 x - 480 + 0$

Add $- 4 x^{3} - 60 x^{2} - 296 x - 480$ and $0$ .

$- 4 x^{3} - 60 x^{2} - 296 x - 480$

Step 3

Find the second derivative of the function.

By the Sum Rule, the derivative of $- 4 x^{3} - 60 x^{2} - 296 x - 480$ with respect to $x$ is $\frac{d}{d x} [- 4 x^{3}] + \frac{d}{d x} [- 60 x^{2}] + \frac{d}{d x} [- 296 x] + \frac{d}{d x} [- 480]$ .

$f'' (x) = \frac{d}{d x} (- 4 x^{3}) + \frac{d}{d x} (- 60 x^{2}) + \frac{d}{d x} (- 296 x) + \frac{d}{d x} (- 480)$

Evaluate $\frac{d}{d x} [- 4 x^{3}]$ .

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 x^{3}$ with respect to $x$ is $- 4 \frac{d}{d x} [x^{3}]$ .

$f'' (x) = - 4 \frac{d}{d x} x^{3} + \frac{d}{d x} (- 60 x^{2}) + \frac{d}{d x} (- 296 x) + \frac{d}{d x} (- 480)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$f'' (x) = - 4 (3 x^{2}) + \frac{d}{d x} (- 60 x^{2}) + \frac{d}{d x} (- 296 x) + \frac{d}{d x} (- 480)$

Multiply $3$ by $- 4$ .

$f'' (x) = - 12 x^{2} + \frac{d}{d x} (- 60 x^{2}) + \frac{d}{d x} (- 296 x) + \frac{d}{d x} (- 480)$

Evaluate $\frac{d}{d x} [- 60 x^{2}]$ .

Since $- 60$ is constant with respect to $x$ , the derivative of $- 60 x^{2}$ with respect to $x$ is $- 60 \frac{d}{d x} [x^{2}]$ .

$f'' (x) = - 12 x^{2} - 60 \frac{d}{d x} x^{2} + \frac{d}{d x} (- 296 x) + \frac{d}{d x} (- 480)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$f'' (x) = - 12 x^{2} - 60 (2 x) + \frac{d}{d x} (- 296 x) + \frac{d}{d x} (- 480)$

Multiply $2$ by $- 60$ .

$f'' (x) = - 12 x^{2} - 120 x + \frac{d}{d x} (- 296 x) + \frac{d}{d x} (- 480)$

Evaluate $\frac{d}{d x} [- 296 x]$ .

Since $- 296$ is constant with respect to $x$ , the derivative of $- 296 x$ with respect to $x$ is $- 296 \frac{d}{d x} [x]$ .

$f'' (x) = - 12 x^{2} - 120 x - 296 \frac{d}{d x} x + \frac{d}{d x} (- 480)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$f'' (x) = - 12 x^{2} - 120 x - 296 \cdot 1 + \frac{d}{d x} (- 480)$

Multiply $- 296$ by $1$ .

$f'' (x) = - 12 x^{2} - 120 x - 296 + \frac{d}{d x} (- 480)$

Differentiate using the Constant Rule.

Since $- 480$ is constant with respect to $x$ , the derivative of $- 480$ with respect to $x$ is $0$ .

$f'' (x) = - 12 x^{2} - 120 x - 296 + 0$

Add $- 12 x^{2} - 120 x - 296$ and $0$ .

$f'' (x) = - 12 x^{2} - 120 x - 296$

Step 4

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$- 4 x^{3} - 60 x^{2} - 296 x - 480 = 0$

Step 5

Find the first derivative.

$f' (x) = \frac{d}{d x} (- x^{4}) + \frac{d}{d x} (- 20 x^{3}) + \frac{d}{d x} (- 148 x^{2}) + \frac{d}{d x} (- 480 x) + \frac{d}{d x} (- 576)$

Evaluate $\frac{d}{d x} [- x^{4}]$ .

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{4}$ with respect to $x$ is $- \frac{d}{d x} [x^{4}]$ .

$f' (x) = - \frac{d}{d x} x^{4} + \frac{d}{d x} (- 20 x^{3}) + \frac{d}{d x} (- 148 x^{2}) + \frac{d}{d x} (- 480 x) + \frac{d}{d x} (- 576)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$f' (x) = - (4 x^{3}) + \frac{d}{d x} (- 20 x^{3}) + \frac{d}{d x} (- 148 x^{2}) + \frac{d}{d x} (- 480 x) + \frac{d}{d x} (- 576)$

Multiply $4$ by $- 1$ .

$f' (x) = - 4 x^{3} + \frac{d}{d x} (- 20 x^{3}) + \frac{d}{d x} (- 148 x^{2}) + \frac{d}{d x} (- 480 x) + \frac{d}{d x} (- 576)$

Evaluate $\frac{d}{d x} [- 20 x^{3}]$ .

Since $- 20$ is constant with respect to $x$ , the derivative of $- 20 x^{3}$ with respect to $x$ is $- 20 \frac{d}{d x} [x^{3}]$ .

$f' (x) = - 4 x^{3} - 20 \frac{d}{d x} x^{3} + \frac{d}{d x} (- 148 x^{2}) + \frac{d}{d x} (- 480 x) + \frac{d}{d x} (- 576)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$f' (x) = - 4 x^{3} - 20 (3 x^{2}) + \frac{d}{d x} (- 148 x^{2}) + \frac{d}{d x} (- 480 x) + \frac{d}{d x} (- 576)$

Multiply $3$ by $- 20$ .

$f' (x) = - 4 x^{3} - 60 x^{2} + \frac{d}{d x} (- 148 x^{2}) + \frac{d}{d x} (- 480 x) + \frac{d}{d x} (- 576)$

Evaluate $\frac{d}{d x} [- 148 x^{2}]$ .

Since $- 148$ is constant with respect to $x$ , the derivative of $- 148 x^{2}$ with respect to $x$ is $- 148 \frac{d}{d x} [x^{2}]$ .

$f' (x) = - 4 x^{3} - 60 x^{2} - 148 \frac{d}{d x} x^{2} + \frac{d}{d x} (- 480 x) + \frac{d}{d x} (- 576)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$f' (x) = - 4 x^{3} - 60 x^{2} - 148 (2 x) + \frac{d}{d x} (- 480 x) + \frac{d}{d x} (- 576)$

Multiply $2$ by $- 148$ .

$f' (x) = - 4 x^{3} - 60 x^{2} - 296 x + \frac{d}{d x} (- 480 x) + \frac{d}{d x} (- 576)$

Evaluate $\frac{d}{d x} [- 480 x]$ .

Since $- 480$ is constant with respect to $x$ , the derivative of $- 480 x$ with respect to $x$ is $- 480 \frac{d}{d x} [x]$ .

$f' (x) = - 4 x^{3} - 60 x^{2} - 296 x - 480 \frac{d}{d x} x + \frac{d}{d x} (- 576)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$f' (x) = - 4 x^{3} - 60 x^{2} - 296 x - 480 \cdot 1 + \frac{d}{d x} (- 576)$

Multiply $- 480$ by $1$ .

$f' (x) = - 4 x^{3} - 60 x^{2} - 296 x - 480 + \frac{d}{d x} (- 576)$

Differentiate using the Constant Rule.

Since $- 576$ is constant with respect to $x$ , the derivative of $- 576$ with respect to $x$ is $0$ .

$f' (x) = - 4 x^{3} - 60 x^{2} - 296 x - 480 + 0$

Add $- 4 x^{3} - 60 x^{2} - 296 x - 480$ and $0$ .

$f' (x) = - 4 x^{3} - 60 x^{2} - 296 x - 480$

The first derivative of $f (x)$ with respect to $x$ is $- 4 x^{3} - 60 x^{2} - 296 x - 480$ .

$- 4 x^{3} - 60 x^{2} - 296 x - 480$

Step 6

Set the first derivative equal to $0$ then solve the equation $- 4 x^{3} - 60 x^{2} - 296 x - 480 = 0$ .

Set the first derivative equal to $0$ .

$- 4 x^{3} - 60 x^{2} - 296 x - 480 = 0$

Factor the left side of the equation.

Factor $4$ out of $- 4 x^{3} - 60 x^{2} - 296 x - 480$ .

Factor $4$ out of $- 4 x^{3}$ .

$4 (- x^{3}) - 60 x^{2} - 296 x - 480 = 0$

Factor $4$ out of $- 60 x^{2}$ .

$4 (- x^{3}) + 4 (- 15 x^{2}) - 296 x - 480 = 0$

Factor $4$ out of $- 296 x$ .

$4 (- x^{3}) + 4 (- 15 x^{2}) + 4 (- 74 x) - 480 = 0$

Factor $4$ out of $- 480$ .

$4 (- x^{3}) + 4 (- 15 x^{2}) + 4 (- 74 x) + 4 (- 120) = 0$

Factor $4$ out of $4 (- x^{3}) + 4 (- 15 x^{2})$ .

$4 (- x^{3} - 15 x^{2}) + 4 (- 74 x) + 4 (- 120) = 0$

Factor $4$ out of $4 (- x^{3} - 15 x^{2}) + 4 (- 74 x)$ .

$4 (- x^{3} - 15 x^{2} - 74 x) + 4 (- 120) = 0$

Factor $4$ out of $4 (- x^{3} - 15 x^{2} - 74 x) + 4 (- 120)$ .

$4 (- x^{3} - 15 x^{2} - 74 x - 120) = 0$

Factor $- x^{3} - 15 x^{2} - 74 x - 120$ using the rational roots test.

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 120, \pm 2, \pm 60, \pm 3, \pm 40, \pm 4, \pm 30, \pm 5, \pm 24, \pm 6, \pm 20, \pm 8, \pm 15, \pm 10, \pm 12 q = \pm 1$

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 120, \pm 2, \pm 60, \pm 3, \pm 40, \pm 4, \pm 30, \pm 5, \pm 24, \pm 6, \pm 20, \pm 8, \pm 15, \pm 10, \pm 12$

Substitute $- 4$ and simplify the expression. In this case, the expression is equal to $0$ so $- 4$ is a root of the polynomial.

Substitute $- 4$ into the polynomial.

$- {(- 4)}^{3} - 15 {(- 4)}^{2} - 74 \cdot - 4 - 120$

Raise $- 4$ to the power of $3$ .

$- - 64 - 15 {(- 4)}^{2} - 74 \cdot - 4 - 120$

Multiply $- 1$ by $- 64$ .

$64 - 15 {(- 4)}^{2} - 74 \cdot - 4 - 120$

Raise $- 4$ to the power of $2$ .

$64 - 15 \cdot 16 - 74 \cdot - 4 - 120$

Multiply $- 15$ by $16$ .

$64 - 240 - 74 \cdot - 4 - 120$

Subtract $240$ from $64$ .

$- 176 - 74 \cdot - 4 - 120$

Multiply $- 74$ by $- 4$ .

$- 176 + 296 - 120$

Add $- 176$ and $296$ .

$120 - 120$

Subtract $120$ from $120$ .

$0$

Since $- 4$ is a known root, divide the polynomial by $x + 4$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{- x^{3} - 15 x^{2} - 74 x - 120}{x + 4}$

Divide $- x^{3} - 15 x^{2} - 74 x - 120$ by $x + 4$ .

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$4$

$x^{3}$

$15 x^{2}$

$74 x$

$120$

Divide the highest order term in the dividend $- x^{3}$ by the highest order term in divisor $x$ .

				-	$x^{2}$
	$x$	+	$4$	-	$x^{3}$	-	$15 x^{2}$	-	$74 x$	-	$120$

Multiply the new quotient term by the divisor.

			-	$x^{2}$
$x$	+	$4$	-	$x^{3}$	-	$15 x^{2}$	-	$74 x$	-	$120$
			-	$x^{3}$	-	$4 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $- x^{3} - 4 x^{2}$

			-	$x^{2}$
$x$	+	$4$	-	$x^{3}$	-	$15 x^{2}$	-	$74 x$	-	$120$
			+	$x^{3}$	+	$4 x^{2}$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

			-	$x^{2}$
$x$	+	$4$	-	$x^{3}$	-	$15 x^{2}$	-	$74 x$	-	$120$
			+	$x^{3}$	+	$4 x^{2}$
					-	$11 x^{2}$

Pull the next terms from the original dividend down into the current dividend.

			-	$x^{2}$
$x$	+	$4$	-	$x^{3}$	-	$15 x^{2}$	-	$74 x$	-	$120$
			+	$x^{3}$	+	$4 x^{2}$
					-	$11 x^{2}$	-	$74 x$

Divide the highest order term in the dividend $- 11 x^{2}$ by the highest order term in divisor $x$ .

			-	$x^{2}$	-	$11 x$
$x$	+	$4$	-	$x^{3}$	-	$15 x^{2}$	-	$74 x$	-	$120$
			+	$x^{3}$	+	$4 x^{2}$
					-	$11 x^{2}$	-	$74 x$

Multiply the new quotient term by the divisor.

			-	$x^{2}$	-	$11 x$
$x$	+	$4$	-	$x^{3}$	-	$15 x^{2}$	-	$74 x$	-	$120$
			+	$x^{3}$	+	$4 x^{2}$
					-	$11 x^{2}$	-	$74 x$
					-	$11 x^{2}$	-	$44 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 11 x^{2} - 44 x$

			-	$x^{2}$	-	$11 x$
$x$	+	$4$	-	$x^{3}$	-	$15 x^{2}$	-	$74 x$	-	$120$
			+	$x^{3}$	+	$4 x^{2}$
					-	$11 x^{2}$	-	$74 x$
					+	$11 x^{2}$	+	$44 x$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

			-	$x^{2}$	-	$11 x$
$x$	+	$4$	-	$x^{3}$	-	$15 x^{2}$	-	$74 x$	-	$120$
			+	$x^{3}$	+	$4 x^{2}$
					-	$11 x^{2}$	-	$74 x$
					+	$11 x^{2}$	+	$44 x$
							-	$30 x$

Pull the next terms from the original dividend down into the current dividend.

			-	$x^{2}$	-	$11 x$
$x$	+	$4$	-	$x^{3}$	-	$15 x^{2}$	-	$74 x$	-	$120$
			+	$x^{3}$	+	$4 x^{2}$
					-	$11 x^{2}$	-	$74 x$
					+	$11 x^{2}$	+	$44 x$
							-	$30 x$	-	$120$

Divide the highest order term in the dividend $- 30 x$ by the highest order term in divisor $x$ .

			-	$x^{2}$	-	$11 x$	-	$30$
$x$	+	$4$	-	$x^{3}$	-	$15 x^{2}$	-	$74 x$	-	$120$
			+	$x^{3}$	+	$4 x^{2}$
					-	$11 x^{2}$	-	$74 x$
					+	$11 x^{2}$	+	$44 x$
							-	$30 x$	-	$120$

Multiply the new quotient term by the divisor.

			-	$x^{2}$	-	$11 x$	-	$30$
$x$	+	$4$	-	$x^{3}$	-	$15 x^{2}$	-	$74 x$	-	$120$
			+	$x^{3}$	+	$4 x^{2}$
					-	$11 x^{2}$	-	$74 x$
					+	$11 x^{2}$	+	$44 x$
							-	$30 x$	-	$120$
							-	$30 x$	-	$120$

The expression needs to be subtracted from the dividend, so change all the signs in $- 30 x - 120$

			-	$x^{2}$	-	$11 x$	-	$30$
$x$	+	$4$	-	$x^{3}$	-	$15 x^{2}$	-	$74 x$	-	$120$
			+	$x^{3}$	+	$4 x^{2}$
					-	$11 x^{2}$	-	$74 x$
					+	$11 x^{2}$	+	$44 x$
							-	$30 x$	-	$120$
							+	$30 x$	+	$120$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

			-	$x^{2}$	-	$11 x$	-	$30$
$x$	+	$4$	-	$x^{3}$	-	$15 x^{2}$	-	$74 x$	-	$120$
			+	$x^{3}$	+	$4 x^{2}$
					-	$11 x^{2}$	-	$74 x$
					+	$11 x^{2}$	+	$44 x$
							-	$30 x$	-	$120$
							+	$30 x$	+	$120$
										$0$

Since the remander is $0$ , the final answer is the quotient.

$- x^{2} - 11 x - 30$

Write $- x^{3} - 15 x^{2} - 74 x - 120$ as a set of factors.

$4 ((x + 4) (- x^{2} - 11 x - 30)) = 0$

Factor.

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot - 30 = 30$ and whose sum is $b = - 11$ .

Factor $- 11$ out of $- 11 x$ .

$4 ((x + 4) (- x^{2} - 11 x - 30)) = 0$

Rewrite $- 11$ as $- 5$ plus $- 6$

$4 ((x + 4) (- x^{2} + (- 5 - 6) x - 30)) = 0$

Apply the distributive property.

$4 ((x + 4) (- x^{2} - 5 x - 6 x - 30)) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$4 ((x + 4) ((- x^{2} - 5 x) - 6 x - 30)) = 0$

Factor out the greatest common factor (GCF) from each group.

$4 ((x + 4) (x (- x - 5) + 6 (- x - 5))) = 0$

Factor the polynomial by factoring out the greatest common factor, $- x - 5$ .

$4 ((x + 4) ((- x - 5) (x + 6))) = 0$

Remove unnecessary parentheses.

$4 ((x + 4) (- x - 5) (x + 6)) = 0$

Remove unnecessary parentheses.

$4 (x + 4) (- x - 5) (x + 6) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x + 4 = 0$

$- x - 5 = 0$

$x + 6 = 0$

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Set $- x - 5$ equal to $0$ and solve for $x$ .

Set $- x - 5$ equal to $0$ .

$- x - 5 = 0$

Solve $- x - 5 = 0$ for $x$ .

Add $5$ to both sides of the equation.

$- x = 5$

Divide each term in $- x = 5$ by $- 1$ and simplify.

Divide each term in $- x = 5$ by $- 1$ .

$\frac{- x}{- 1} = \frac{5}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{5}{- 1}$

Divide $x$ by $1$ .

$x = \frac{5}{- 1}$

Simplify the right side.

Divide $5$ by $- 1$ .

$x = - 5$

Set $x + 6$ equal to $0$ and solve for $x$ .

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

The final solution is all the values that make $4 (x + 4) (- x - 5) (x + 6) = 0$ true.

$x = - 4, - 5, - 6$

Step 7

Find the values where the derivative is undefined.

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Step 8

Critical points to evaluate.

$x = - 4, - 5, - 6$

Step 9

Evaluate the second derivative at $x = - 4$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 12 {(- 4)}^{2} - 120 \cdot - 4 - 296$

Step 10

Evaluate the second derivative.

Simplify each term.

Raise $- 4$ to the power of $2$ .

$- 12 \cdot 16 - 120 \cdot - 4 - 296$

Multiply $- 12$ by $16$ .

$- 192 - 120 \cdot - 4 - 296$

Multiply $- 120$ by $- 4$ .

$- 192 + 480 - 296$

Simplify by adding and subtracting.

Add $- 192$ and $480$ .

$288 - 296$

Subtract $296$ from $288$ .

$- 8$

Step 11

$x = - 4$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = - 4$ is a local maximum

Step 12

Find the y-value when $x = - 4$ .

Replace the variable $x$ with $- 4$ in the expression.

$f (- 4) = - {(- 4)}^{4} - 20 {(- 4)}^{3} - 148 {(- 4)}^{2} - 480 \cdot - 4 - 576$

Simplify the result.

Simplify each term.

Raise $- 4$ to the power of $4$ .

$f (- 4) = - 1 \cdot 256 - 20 {(- 4)}^{3} - 148 {(- 4)}^{2} - 480 \cdot - 4 - 576$

Multiply $- 1$ by $256$ .

$f (- 4) = - 256 - 20 {(- 4)}^{3} - 148 {(- 4)}^{2} - 480 \cdot - 4 - 576$

Raise $- 4$ to the power of $3$ .

$f (- 4) = - 256 - 20 \cdot - 64 - 148 {(- 4)}^{2} - 480 \cdot - 4 - 576$

Multiply $- 20$ by $- 64$ .

$f (- 4) = - 256 + 1280 - 148 {(- 4)}^{2} - 480 \cdot - 4 - 576$

Raise $- 4$ to the power of $2$ .

$f (- 4) = - 256 + 1280 - 148 \cdot 16 - 480 \cdot - 4 - 576$

Multiply $- 148$ by $16$ .

$f (- 4) = - 256 + 1280 - 2368 - 480 \cdot - 4 - 576$

Multiply $- 480$ by $- 4$ .

$f (- 4) = - 256 + 1280 - 2368 + 1920 - 576$

Simplify by adding and subtracting.

Add $- 256$ and $1280$ .

$f (- 4) = 1024 - 2368 + 1920 - 576$

Subtract $2368$ from $1024$ .

$f (- 4) = - 1344 + 1920 - 576$

Add $- 1344$ and $1920$ .

$f (- 4) = 576 - 576$

Subtract $576$ from $576$ .

$f (- 4) = 0$

The final answer is $0$ .

$y = 0$

Step 13

Evaluate the second derivative at $x = - 5$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 12 {(- 5)}^{2} - 120 \cdot - 5 - 296$

Step 14

Evaluate the second derivative.

Simplify each term.

Raise $- 5$ to the power of $2$ .

$- 12 \cdot 25 - 120 \cdot - 5 - 296$

Multiply $- 12$ by $25$ .

$- 300 - 120 \cdot - 5 - 296$

Multiply $- 120$ by $- 5$ .

$- 300 + 600 - 296$

Simplify by adding and subtracting.

Add $- 300$ and $600$ .

$300 - 296$

Subtract $296$ from $300$ .

$4$

Step 15

$x = - 5$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = - 5$ is a local minimum

Step 16

Find the y-value when $x = - 5$ .

Replace the variable $x$ with $- 5$ in the expression.

$f (- 5) = - {(- 5)}^{4} - 20 {(- 5)}^{3} - 148 {(- 5)}^{2} - 480 \cdot - 5 - 576$

Simplify the result.

Simplify each term.

Raise $- 5$ to the power of $4$ .

$f (- 5) = - 1 \cdot 625 - 20 {(- 5)}^{3} - 148 {(- 5)}^{2} - 480 \cdot - 5 - 576$

Multiply $- 1$ by $625$ .

$f (- 5) = - 625 - 20 {(- 5)}^{3} - 148 {(- 5)}^{2} - 480 \cdot - 5 - 576$

Raise $- 5$ to the power of $3$ .

$f (- 5) = - 625 - 20 \cdot - 125 - 148 {(- 5)}^{2} - 480 \cdot - 5 - 576$

Multiply $- 20$ by $- 125$ .

$f (- 5) = - 625 + 2500 - 148 {(- 5)}^{2} - 480 \cdot - 5 - 576$

Raise $- 5$ to the power of $2$ .

$f (- 5) = - 625 + 2500 - 148 \cdot 25 - 480 \cdot - 5 - 576$

Multiply $- 148$ by $25$ .

$f (- 5) = - 625 + 2500 - 3700 - 480 \cdot - 5 - 576$

Multiply $- 480$ by $- 5$ .

$f (- 5) = - 625 + 2500 - 3700 + 2400 - 576$

Simplify by adding and subtracting.

Add $- 625$ and $2500$ .

$f (- 5) = 1875 - 3700 + 2400 - 576$

Subtract $3700$ from $1875$ .

$f (- 5) = - 1825 + 2400 - 576$

Add $- 1825$ and $2400$ .

$f (- 5) = 575 - 576$

Subtract $576$ from $575$ .

$f (- 5) = - 1$

The final answer is $- 1$ .

$y = - 1$

Step 17

Evaluate the second derivative at $x = - 6$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 12 {(- 6)}^{2} - 120 \cdot - 6 - 296$

Step 18

Evaluate the second derivative.

Simplify each term.

Raise $- 6$ to the power of $2$ .

$- 12 \cdot 36 - 120 \cdot - 6 - 296$

Multiply $- 12$ by $36$ .

$- 432 - 120 \cdot - 6 - 296$

Multiply $- 120$ by $- 6$ .

$- 432 + 720 - 296$

Simplify by adding and subtracting.

Add $- 432$ and $720$ .

$288 - 296$

Subtract $296$ from $288$ .

$- 8$

Step 19

$x = - 6$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = - 6$ is a local maximum

Step 20

Find the y-value when $x = - 6$ .

Replace the variable $x$ with $- 6$ in the expression.

$f (- 6) = - {(- 6)}^{4} - 20 {(- 6)}^{3} - 148 {(- 6)}^{2} - 480 \cdot - 6 - 576$

Simplify the result.

Simplify each term.

Raise $- 6$ to the power of $4$ .

$f (- 6) = - 1 \cdot 1296 - 20 {(- 6)}^{3} - 148 {(- 6)}^{2} - 480 \cdot - 6 - 576$

Multiply $- 1$ by $1296$ .

$f (- 6) = - 1296 - 20 {(- 6)}^{3} - 148 {(- 6)}^{2} - 480 \cdot - 6 - 576$

Raise $- 6$ to the power of $3$ .

$f (- 6) = - 1296 - 20 \cdot - 216 - 148 {(- 6)}^{2} - 480 \cdot - 6 - 576$

Multiply $- 20$ by $- 216$ .

$f (- 6) = - 1296 + 4320 - 148 {(- 6)}^{2} - 480 \cdot - 6 - 576$

Raise $- 6$ to the power of $2$ .

$f (- 6) = - 1296 + 4320 - 148 \cdot 36 - 480 \cdot - 6 - 576$

Multiply $- 148$ by $36$ .

$f (- 6) = - 1296 + 4320 - 5328 - 480 \cdot - 6 - 576$

Multiply $- 480$ by $- 6$ .

$f (- 6) = - 1296 + 4320 - 5328 + 2880 - 576$

Simplify by adding and subtracting.

Add $- 1296$ and $4320$ .

$f (- 6) = 3024 - 5328 + 2880 - 576$

Subtract $5328$ from $3024$ .

$f (- 6) = - 2304 + 2880 - 576$

Add $- 2304$ and $2880$ .

$f (- 6) = 576 - 576$

Subtract $576$ from $576$ .

$f (- 6) = 0$

The final answer is $0$ .

$y = 0$

Step 21

These are the local extrema for $f (x) = - x^{4} - 20 x^{3} - 148 x^{2} - 480 x - 576$ .

$(- 4, 0)$ is a local maxima

$(- 5, - 1)$ is a local minima

$(- 6, 0)$ is a local maxima

Step 22

Find the Local Maxima and Minima y=-x^4-20x^3-148x^2-480x-576

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