Find the Local Maxima and Minima x^3+x^2-x-1

$x^{3} + x^{2} - x - 1$

Write $x^{3} + x^{2} - x - 1$ as a function.

$f (x) = x^{3} + x^{2} - x - 1$

Find the first derivative of the function.

Differentiate.

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

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

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

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

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

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

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

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

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

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

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

Multiply $- 1$ by $1$ .

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

Differentiate using the Constant Rule.

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$3 x^{2} + 2 x - 1 + 0$

Add $3 x^{2} + 2 x - 1$ and $0$ .

$3 x^{2} + 2 x - 1$

Find the second derivative of the function.

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

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

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

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

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

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

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

Multiply $2$ by $3$ .

$f'' (x) = 6 x + \frac{d}{d x} (2 x) + \frac{d}{d x} (- 1)$

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

$f'' (x) = 6 x + 2 \frac{d}{d x} (x) + \frac{d}{d x} (- 1)$

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

$f'' (x) = 6 x + 2 \cdot 1 + \frac{d}{d x} (- 1)$

Multiply $2$ by $1$ .

$f'' (x) = 6 x + 2 + \frac{d}{d x} (- 1)$

Differentiate using the Constant Rule.

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$f'' (x) = 6 x + 2 + 0$

Add $6 x + 2$ and $0$ .

$f'' (x) = 6 x + 2$

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

$3 x^{2} + 2 x - 1 = 0$

Find the first derivative.

Differentiate.

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

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

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

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

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

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

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

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

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

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

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

Multiply $- 1$ by $1$ .

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

Differentiate using the Constant Rule.

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$f' (x) = 3 x^{2} + 2 x - 1 + 0$

Add $3 x^{2} + 2 x - 1$ and $0$ .

$f' (x) = 3 x^{2} + 2 x - 1$

The first derivative of $f (x)$ with respect to $x$ is $3 x^{2} + 2 x - 1$ .

$3 x^{2} + 2 x - 1$

Set the first derivative equal to $0$ then solve the equation $3 x^{2} + 2 x - 1 = 0$ .

Set the first derivative equal to $0$ .

$3 x^{2} + 2 x - 1 = 0$

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 $a \cdot c = 3 \cdot - 1 = - 3$ and whose sum is $b = 2$ .

Factor $2$ out of $2 x$ .

$3 x^{2} + 2 (x) - 1 = 0$

Rewrite $2$ as $- 1$ plus $3$

$3 x^{2} + (- 1 + 3) x - 1 = 0$

Apply the distributive property.

$3 x^{2} - 1 x + 3 x - 1 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(3 x^{2} - 1 x) + 3 x - 1 = 0$

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

$x (3 x - 1) + 1 (3 x - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $3 x - 1$ .

$(3 x - 1) (x + 1) = 0$

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

$3 x - 1 = 0$

$x + 1 = 0$

Set $3 x - 1$ equal to $0$ and solve for $x$ .

Set $3 x - 1$ equal to $0$ .

$3 x - 1 = 0$

Solve $3 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$3 x = 1$

Divide each term in $3 x = 1$ by $3$ and simplify.

Divide each term in $3 x = 1$ by $3$ .

$\frac{3 x}{3} = \frac{1}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{1}{3}$

Divide $x$ by $1$ .

$x = \frac{1}{3}$

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The final solution is all the values that make $(3 x - 1) (x + 1) = 0$ true.

$x = \frac{1}{3}, - 1$

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.

Critical points to evaluate.

$x = \frac{1}{3}, - 1$

Evaluate the second derivative at $x = \frac{1}{3}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$6 (\frac{1}{3}) + 2$

Evaluate the second derivative.

Cancel the common factor of $3$ .

Factor $3$ out of $6$ .

$3 (2) \frac{1}{3} + 2$

Cancel the common factor.

$3 \cdot 2 \frac{1}{3} + 2$

Rewrite the expression.

$2 + 2$

Add $2$ and $2$ .

$4$

$x = \frac{1}{3}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \frac{1}{3}$ is a local minimum

Find the y-value when $x = \frac{1}{3}$ .

Replace the variable $x$ with $\frac{1}{3}$ in the expression.

$f (\frac{1}{3}) = {(\frac{1}{3})}^{3} + {(\frac{1}{3})}^{2} - (\frac{1}{3}) - 1$

Simplify the result.

Simplify each term.

Apply the product rule to $\frac{1}{3}$ .

$f (\frac{1}{3}) = \frac{1^{3}}{3^{3}} + {(\frac{1}{3})}^{2} - (\frac{1}{3}) - 1$

One to any power is one.

$f (\frac{1}{3}) = \frac{1}{3^{3}} + {(\frac{1}{3})}^{2} - (\frac{1}{3}) - 1$

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

$f (\frac{1}{3}) = \frac{1}{27} + {(\frac{1}{3})}^{2} - (\frac{1}{3}) - 1$

Apply the product rule to $\frac{1}{3}$ .

$f (\frac{1}{3}) = \frac{1}{27} + \frac{1^{2}}{3^{2}} - (\frac{1}{3}) - 1$

One to any power is one.

$f (\frac{1}{3}) = \frac{1}{27} + \frac{1}{3^{2}} - (\frac{1}{3}) - 1$

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

$f (\frac{1}{3}) = \frac{1}{27} + \frac{1}{9} - \frac{1}{3} - 1$

Find the common denominator.

Multiply $\frac{1}{9}$ by $\frac{3}{3}$ .

$f (\frac{1}{3}) = \frac{1}{27} + \frac{1}{9} \cdot \frac{3}{3} - \frac{1}{3} - 1$

Multiply $\frac{1}{9}$ by $\frac{3}{3}$ .

$f (\frac{1}{3}) = \frac{1}{27} + \frac{3}{9 \cdot 3} - \frac{1}{3} - 1$

Multiply $\frac{1}{3}$ by $\frac{9}{9}$ .

$f (\frac{1}{3}) = \frac{1}{27} + \frac{3}{9 \cdot 3} - (\frac{1}{3} \cdot \frac{9}{9}) - 1$

Multiply $\frac{1}{3}$ by $\frac{9}{9}$ .

$f (\frac{1}{3}) = \frac{1}{27} + \frac{3}{9 \cdot 3} - \frac{9}{3 \cdot 9} - 1$

Write $- 1$ as a fraction with denominator $1$ .

$f (\frac{1}{3}) = \frac{1}{27} + \frac{3}{9 \cdot 3} - \frac{9}{3 \cdot 9} + \frac{- 1}{1}$

Multiply $\frac{- 1}{1}$ by $\frac{27}{27}$ .

$f (\frac{1}{3}) = \frac{1}{27} + \frac{3}{9 \cdot 3} - \frac{9}{3 \cdot 9} + \frac{- 1}{1} \cdot \frac{27}{27}$

Multiply $\frac{- 1}{1}$ by $\frac{27}{27}$ .

$f (\frac{1}{3}) = \frac{1}{27} + \frac{3}{9 \cdot 3} - \frac{9}{3 \cdot 9} + \frac{- 1 \cdot 27}{27}$

Reorder the factors of $9 \cdot 3$ .

$f (\frac{1}{3}) = \frac{1}{27} + \frac{3}{3 \cdot 9} - \frac{9}{3 \cdot 9} + \frac{- 1 \cdot 27}{27}$

Multiply $3$ by $9$ .

$f (\frac{1}{3}) = \frac{1}{27} + \frac{3}{27} - \frac{9}{3 \cdot 9} + \frac{- 1 \cdot 27}{27}$

Multiply $3$ by $9$ .

$f (\frac{1}{3}) = \frac{1}{27} + \frac{3}{27} - \frac{9}{27} + \frac{- 1 \cdot 27}{27}$

Combine the numerators over the common denominator.

$f (\frac{1}{3}) = \frac{1 + 3 - 9 - 1 \cdot 27}{27}$

Simplify the expression.

Multiply $- 1$ by $27$ .

$f (\frac{1}{3}) = \frac{1 + 3 - 9 - 27}{27}$

Add $1$ and $3$ .

$f (\frac{1}{3}) = \frac{4 - 9 - 27}{27}$

Subtract $9$ from $4$ .

$f (\frac{1}{3}) = \frac{- 5 - 27}{27}$

Subtract $27$ from $- 5$ .

$f (\frac{1}{3}) = \frac{- 32}{27}$

Move the negative in front of the fraction.

$f (\frac{1}{3}) = - \frac{32}{27}$

The final answer is $- \frac{32}{27}$ .

$y = - \frac{32}{27}$

Evaluate the second derivative at $x = - 1$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$6 (- 1) + 2$

Evaluate the second derivative.

Multiply $6$ by $- 1$ .

$- 6 + 2$

Add $- 6$ and $2$ .

$- 4$

$x = - 1$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = - 1$ is a local maximum

Find the y-value when $x = - 1$ .

Replace the variable $x$ with $- 1$ in the expression.

$f (- 1) = {(- 1)}^{3} + {(- 1)}^{2} - (- 1) - 1$

Simplify the result.

Simplify each term.

Raise $- 1$ to the power of $3$ .

$f (- 1) = - 1 + {(- 1)}^{2} - (- 1) - 1$

Raise $- 1$ to the power of $2$ .

$f (- 1) = - 1 + 1 - (- 1) - 1$

Multiply $- 1$ by $- 1$ .

$f (- 1) = - 1 + 1 + 1 - 1$

Simplify by adding and subtracting.

Add $- 1$ and $1$ .

$f (- 1) = 0 + 1 - 1$

Add $0$ and $1$ .

$f (- 1) = 1 - 1$

Subtract $1$ from $1$ .

$f (- 1) = 0$

The final answer is $0$ .

$y = 0$

These are the local extrema for $f (x) = x^{3} + x^{2} - x - 1$ .

$(\frac{1}{3}, - \frac{32}{27})$ is a local minima

$(- 1, 0)$ is a local maxima

Find the Local Maxima and Minima x^3+x^2-x-1

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