Find the Local Maxima and Minima h(x)=1/8x^3-x^2

$h (x) = \frac{1}{8} x^{3} - x^{2}$

Step 1

Find the first derivative of the function.

By the Sum Rule, the derivative of $\frac{1}{8} x^{3} - x^{2}$ with respect to $x$ is $\frac{d}{d x} [\frac{1}{8} x^{3}] + \frac{d}{d x} [- x^{2}]$ .

$\frac{d}{d x} [\frac{1}{8} x^{3}] + \frac{d}{d x} [- x^{2}]$

Evaluate $\frac{d}{d x} [\frac{1}{8} x^{3}]$ .

Since $\frac{1}{8}$ is constant with respect to $x$ , the derivative of $\frac{1}{8} x^{3}$ with respect to $x$ is $\frac{1}{8} \frac{d}{d x} [x^{3}]$ .

$\frac{1}{8} \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- x^{2}]$

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

$\frac{1}{8} (3 x^{2}) + \frac{d}{d x} [- x^{2}]$

Combine $3$ and $\frac{1}{8}$ .

$\frac{3}{8} x^{2} + \frac{d}{d x} [- x^{2}]$

Combine $\frac{3}{8}$ and $x^{2}$ .

$\frac{3 x^{2}}{8} + \frac{d}{d x} [- x^{2}]$

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

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

$\frac{3 x^{2}}{8} - \frac{d}{d x} [x^{2}]$

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

$\frac{3 x^{2}}{8} - (2 x)$

Multiply $2$ by $- 1$ .

$\frac{3 x^{2}}{8} - 2 x$

Step 2

Find the second derivative of the function.

By the Sum Rule, the derivative of $\frac{3 x^{2}}{8} - 2 x$ with respect to $x$ is $\frac{d}{d x} [\frac{3 x^{2}}{8}] + \frac{d}{d x} [- 2 x]$ .

$h'' (x) = \frac{d}{d x} (\frac{3 x^{2}}{8}) + \frac{d}{d x} (- 2 x)$

Evaluate $\frac{d}{d x} [\frac{3 x^{2}}{8}]$ .

Since $\frac{3}{8}$ is constant with respect to $x$ , the derivative of $\frac{3 x^{2}}{8}$ with respect to $x$ is $\frac{3}{8} \frac{d}{d x} [x^{2}]$ .

$h'' (x) = \frac{3}{8} \cdot \frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 2 x)$

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

$h'' (x) = \frac{3}{8} \cdot (2 x) + \frac{d}{d x} (- 2 x)$

Combine $2$ and $\frac{3}{8}$ .

$h'' (x) = \frac{2 \cdot 3}{8} x + \frac{d}{d x} (- 2 x)$

Multiply $2$ by $3$ .

$h'' (x) = \frac{6}{8} x + \frac{d}{d x} (- 2 x)$

Combine $\frac{6}{8}$ and $x$ .

$h'' (x) = \frac{6 x}{8} + \frac{d}{d x} (- 2 x)$

Cancel the common factor of $6$ and $8$ .

Factor $2$ out of $6 x$ .

$h'' (x) = \frac{2 (3 x)}{8} + \frac{d}{d x} (- 2 x)$

Cancel the common factors.

Factor $2$ out of $8$ .

$h'' (x) = \frac{2 (3 x)}{2 (4)} + \frac{d}{d x} (- 2 x)$

Cancel the common factor.

$h'' (x) = \frac{2 (3 x)}{2 \cdot 4} + \frac{d}{d x} (- 2 x)$

Rewrite the expression.

$h'' (x) = \frac{3 x}{4} + \frac{d}{d x} (- 2 x)$

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

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

$h'' (x) = \frac{3 x}{4} - 2 \frac{d}{d x} x$

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

$h'' (x) = \frac{3 x}{4} - 2 \cdot 1$

Multiply $- 2$ by $1$ .

$h'' (x) = \frac{3 x}{4} - 2$

Step 3

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

$\frac{3 x^{2}}{8} - 2 x = 0$

Step 4

Find the first derivative.

By the Sum Rule, the derivative of $\frac{1}{8} x^{3} - x^{2}$ with respect to $x$ is $\frac{d}{d x} [\frac{1}{8} x^{3}] + \frac{d}{d x} [- x^{2}]$ .

$f' (x) = \frac{d}{d x} (\frac{1}{8} x^{3}) + \frac{d}{d x} (- x^{2})$

Evaluate $\frac{d}{d x} [\frac{1}{8} x^{3}]$ .

Since $\frac{1}{8}$ is constant with respect to $x$ , the derivative of $\frac{1}{8} x^{3}$ with respect to $x$ is $\frac{1}{8} \frac{d}{d x} [x^{3}]$ .

$f' (x) = \frac{1}{8} \cdot \frac{d}{d x} (x^{3}) + \frac{d}{d x} (- x^{2})$

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

$f' (x) = \frac{1}{8} \cdot (3 x^{2}) + \frac{d}{d x} (- x^{2})$

Combine $3$ and $\frac{1}{8}$ .

$f' (x) = \frac{3}{8} x^{2} + \frac{d}{d x} (- x^{2})$

Combine $\frac{3}{8}$ and $x^{2}$ .

$f' (x) = \frac{3 x^{2}}{8} + \frac{d}{d x} (- x^{2})$

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

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

$f' (x) = \frac{3 x^{2}}{8} - \frac{d}{d x} x^{2}$

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

$f' (x) = \frac{3 x^{2}}{8} - (2 x)$

Multiply $2$ by $- 1$ .

$f' (x) = \frac{3 x^{2}}{8} - 2 x$

The first derivative of $h (x)$ with respect to $x$ is $\frac{3 x^{2}}{8} - 2 x$ .

$\frac{3 x^{2}}{8} - 2 x$

Step 5

Set the first derivative equal to $0$ then solve the equation $\frac{3 x^{2}}{8} - 2 x = 0$ .

Set the first derivative equal to $0$ .

$\frac{3 x^{2}}{8} - 2 x = 0$

Multiply each term in $\frac{3 x^{2}}{8} - 2 x = 0$ by $8$ to eliminate the fractions.

Multiply each term in $\frac{3 x^{2}}{8} - 2 x = 0$ by $8$ .

$\frac{3 x^{2}}{8} \cdot 8 - 2 x \cdot 8 = 0 \cdot 8$

Simplify the left side.

Simplify each term.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{3 x^{2}}{8} \cdot 8 - 2 x \cdot 8 = 0 \cdot 8$

Rewrite the expression.

$3 x^{2} - 2 x \cdot 8 = 0 \cdot 8$

Multiply $8$ by $- 2$ .

$3 x^{2} - 16 x = 0 \cdot 8$

Simplify the right side.

Multiply $0$ by $8$ .

$3 x^{2} - 16 x = 0$

Factor $x$ out of $3 x^{2} - 16 x$ .

Factor $x$ out of $3 x^{2}$ .

$x (3 x) - 16 x = 0$

Factor $x$ out of $- 16 x$ .

$x (3 x) + x \cdot - 16 = 0$

Factor $x$ out of $x (3 x) + x \cdot - 16$ .

$x (3 x - 16) = 0$

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

$x = 0$

$3 x - 16 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $3 x - 16$ equal to $0$ and solve for $x$ .

Set $3 x - 16$ equal to $0$ .

$3 x - 16 = 0$

Solve $3 x - 16 = 0$ for $x$ .

Add $16$ to both sides of the equation.

$3 x = 16$

Divide each term in $3 x = 16$ by $3$ and simplify.

Divide each term in $3 x = 16$ by $3$ .

$\frac{3 x}{3} = \frac{16}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{16}{3}$

Divide $x$ by $1$ .

$x = \frac{16}{3}$

The final solution is all the values that make $x (3 x - 16) = 0$ true.

$x = 0, \frac{16}{3}$

Step 6

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 7

Critical points to evaluate.

$x = 0, \frac{16}{3}$

Step 8

Evaluate the second derivative at $x = 0$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{3 (0)}{4} - 2$

Step 9

Evaluate the second derivative.

Multiply $3$ by $0$ .

$\frac{0}{4} - 2$

Divide $0$ by $4$ .

$0 - 2$

Subtract $2$ from $0$ .

$- 2$

Step 10

$x = 0$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = 0$ is a local maximum

Step 11

Find the y-value when $x = 0$ .

Replace the variable $x$ with $0$ in the expression.

$h (0) = \frac{1}{8} \cdot {(0)}^{3} - {(0)}^{2}$

Simplify the result.

Simplify each term.

Raising $0$ to any positive power yields $0$ .

$h (0) = \frac{1}{8} \cdot 0 - {(0)}^{2}$

Multiply $\frac{1}{8}$ by $0$ .

$h (0) = 0 - {(0)}^{2}$

Raising $0$ to any positive power yields $0$ .

$h (0) = 0 - 0$

Multiply $- 1$ by $0$ .

$h (0) = 0 + 0$

Add $0$ and $0$ .

$h (0) = 0$

The final answer is $0$ .

$y = 0$

Step 12

Evaluate the second derivative at $x = \frac{16}{3}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{3 (\frac{16}{3})}{4} - 2$

Step 13

Evaluate the second derivative.

Simplify each term.

Combine $3$ and $\frac{16}{3}$ .

$\frac{\frac{3 \cdot 16}{3}}{4} - 2$

Multiply $3$ by $16$ .

$\frac{\frac{48}{3}}{4} - 2$

Divide $48$ by $3$ .

$\frac{16}{4} - 2$

Divide $16$ by $4$ .

$4 - 2$

Subtract $2$ from $4$ .

$2$

Step 14

$x = \frac{16}{3}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \frac{16}{3}$ is a local minimum

Step 15

Find the y-value when $x = \frac{16}{3}$ .

Replace the variable $x$ with $\frac{16}{3}$ in the expression.

$h (\frac{16}{3}) = \frac{1}{8} \cdot {(\frac{16}{3})}^{3} - {(\frac{16}{3})}^{2}$

Simplify the result.

Simplify each term.

Apply the product rule to $\frac{16}{3}$ .

$h (\frac{16}{3}) = \frac{1}{8} \cdot \frac{16^{3}}{3^{3}} - {(\frac{16}{3})}^{2}$

Combine.

$h (\frac{16}{3}) = \frac{1 \cdot 16^{3}}{8 \cdot 3^{3}} - {(\frac{16}{3})}^{2}$

Multiply $16^{3}$ by $1$ .

$h (\frac{16}{3}) = \frac{16^{3}}{8 \cdot 3^{3}} - {(\frac{16}{3})}^{2}$

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

$h (\frac{16}{3}) = \frac{16^{3}}{8 \cdot 27} - {(\frac{16}{3})}^{2}$

Raise $16$ to the power of $3$ .

$h (\frac{16}{3}) = \frac{4096}{8 \cdot 27} - {(\frac{16}{3})}^{2}$

Multiply $8$ by $27$ .

$h (\frac{16}{3}) = \frac{4096}{216} - {(\frac{16}{3})}^{2}$

Cancel the common factor of $4096$ and $216$ .

Factor $8$ out of $4096$ .

$h (\frac{16}{3}) = \frac{8 (512)}{216} - {(\frac{16}{3})}^{2}$

Cancel the common factors.

Factor $8$ out of $216$ .

$h (\frac{16}{3}) = \frac{8 \cdot 512}{8 \cdot 27} - {(\frac{16}{3})}^{2}$

Cancel the common factor.

$h (\frac{16}{3}) = \frac{8 \cdot 512}{8 \cdot 27} - {(\frac{16}{3})}^{2}$

Rewrite the expression.

$h (\frac{16}{3}) = \frac{512}{27} - {(\frac{16}{3})}^{2}$

Apply the product rule to $\frac{16}{3}$ .

$h (\frac{16}{3}) = \frac{512}{27} - \frac{16^{2}}{3^{2}}$

Raise $16$ to the power of $2$ .

$h (\frac{16}{3}) = \frac{512}{27} - \frac{256}{3^{2}}$

Raise $3$ to the power of $2$ .

$h (\frac{16}{3}) = \frac{512}{27} - \frac{256}{9}$

To write $- \frac{256}{9}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$h (\frac{16}{3}) = \frac{512}{27} - \frac{256}{9} \cdot \frac{3}{3}$

Write each expression with a common denominator of $27$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{256}{9}$ by $\frac{3}{3}$ .

$h (\frac{16}{3}) = \frac{512}{27} - \frac{256 \cdot 3}{9 \cdot 3}$

Multiply $9$ by $3$ .

$h (\frac{16}{3}) = \frac{512}{27} - \frac{256 \cdot 3}{27}$

Combine the numerators over the common denominator.

$h (\frac{16}{3}) = \frac{512 - 256 \cdot 3}{27}$

Simplify the numerator.

Multiply $- 256$ by $3$ .

$h (\frac{16}{3}) = \frac{512 - 768}{27}$

Subtract $768$ from $512$ .

$h (\frac{16}{3}) = \frac{- 256}{27}$

Move the negative in front of the fraction.

$h (\frac{16}{3}) = - \frac{256}{27}$

The final answer is $- \frac{256}{27}$ .

$y = - \frac{256}{27}$

Step 16

These are the local extrema for $h (x) = \frac{1}{8} x^{3} - x^{2}$ .

$(0, 0)$ is a local maxima

$(\frac{16}{3}, - \frac{256}{27})$ is a local minima

Step 17

Find the Local Maxima and Minima h(x)=1/8x^3-x^2

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