Calculus Examples

$f (x) = x^{4} + 2 x^{2} - 8 x$

Find the derivative.

Tap for more steps...

Differentiate.

Tap for more steps...

By the Sum Rule, the derivative of $x^{4} + 2 x^{2} - 8 x$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [- 8 x]$ .

$\frac{d}{d x} [x^{4}] + \frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [- 8 x]$

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

$4 x^{3} + \frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [- 8 x]$

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

Tap for more steps...

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

$4 x^{3} + 2 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 8 x]$

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

$4 x^{3} + 2 (2 x) + \frac{d}{d x} [- 8 x]$

Multiply $2$ by $2$ .

$4 x^{3} + 4 x + \frac{d}{d x} [- 8 x]$

Evaluate $\frac{d}{d x} [- 8 x]$ .

Tap for more steps...

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

$4 x^{3} + 4 x - 8 \frac{d}{d x} [x]$

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

$4 x^{3} + 4 x - 8 \cdot 1$

Multiply $- 8$ by $1$ .

$4 x^{3} + 4 x - 8$

Set the derivative equal to $0$ .

$4 x^{3} + 4 x - 8 = 0$

Solve for $x$ .

Tap for more steps...

Factor the left side of the equation.

Tap for more steps...

Factor $4$ out of $4 x^{3} + 4 x - 8$ .

Tap for more steps...

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

$4 (x^{3}) + 4 x - 8 = 0$

Factor $4$ out of $4 x$ .

$4 (x^{3}) + 4 (x) - 8 = 0$

Factor $4$ out of $- 8$ .

$4 (x^{3}) + 4 x + 4 \cdot - 2 = 0$

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

$4 (x^{3} + x) + 4 \cdot - 2 = 0$

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

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

Factor.

Tap for more steps...

Factor $x^{3} + x - 2$ using the rational roots test.

Tap for more steps...

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.

$p = \pm 1, \pm 2 q = \pm 1$

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

$\pm 1, \pm 2$

Substitute $1$ and simplify the expression. In this case, the expression is equal to $0$ so $1$ is a root of the polynomial.

Tap for more steps...

Substitute $1$ into the polynomial.

$1^{3} + 1 - 2$

One to any power is one.

$1 + 1 - 2$

Add $1$ and $1$ .

$2 - 2$

Subtract $2$ from $2$ .

$0$

Since $1$ is a known root, divide the polynomial by $x - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

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

Divide $x^{3} + x - 2$ by $x - 1$ .

$x^{2} + x + 2$

Write $x^{3} + x - 2$ as a set of factors.

$4 ((x - 1) (x^{2} + x + 2)) = 0$

Remove unnecessary parentheses.

$4 (x - 1) (x^{2} + x + 2) = 0$

Divide both sides of the equation by $4$ . Dividing $0$ by any non-zero number is $0$ .

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

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

Tap for more steps...

Set the factor equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Set $x^{2} + x + 2$ equal to $0$ and solve for $x$ .

Tap for more steps...

Set the factor equal to $0$ .

$x^{2} + x + 2 = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = 1$ , and $c = 2$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (1 \cdot 2)}}{2 \cdot 1}$

Simplify.

Tap for more steps...

Simplify the numerator.

Tap for more steps...

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot (1 \cdot 2)}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 2}}{2 \cdot 1}$

Multiply $- 4$ by $2$ .

$x = \frac{- 1 \pm \sqrt{1 - 8}}{2 \cdot 1}$

Subtract $8$ from $1$ .

$x = \frac{- 1 \pm \sqrt{- 7}}{2 \cdot 1}$

Rewrite $- 7$ as $- 1 (7)$ .

$x = \frac{- 1 \pm \sqrt{- 1 \cdot 7}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (7)}$ as $\sqrt{- 1} \cdot \sqrt{7}$ .

$x = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{7}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 1 \pm i \sqrt{7}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm i \sqrt{7}}{2}$

Factor $- 1$ out of $- 1 \pm i \sqrt{7}$ .

$x = \frac{- 11 \pm i \sqrt{7}}{2}$

Multiply $- 1$ by $- 1$ .

$x = \frac{1 - 1 \pm i \sqrt{7}}{2}$

Multiply $- 1 \pm i \sqrt{7}$ by $1$ .

$x = \frac{- 1 \pm i \sqrt{7}}{2}$

Simplify the expression to solve for the $+$ portion of the $\pm$ .

Tap for more steps...

Simplify the numerator.

Tap for more steps...

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot (1 \cdot 2)}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 2}}{2 \cdot 1}$

Multiply $- 4$ by $2$ .

$x = \frac{- 1 \pm \sqrt{1 - 8}}{2 \cdot 1}$

Subtract $8$ from $1$ .

$x = \frac{- 1 \pm \sqrt{- 7}}{2 \cdot 1}$

Rewrite $- 7$ as $- 1 (7)$ .

$x = \frac{- 1 \pm \sqrt{- 1 \cdot 7}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (7)}$ as $\sqrt{- 1} \cdot \sqrt{7}$ .

$x = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{7}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 1 \pm i \sqrt{7}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm i \sqrt{7}}{2}$

Factor $- 1$ out of $- 1 \pm i \sqrt{7}$ .

$x = \frac{- 11 \pm i \sqrt{7}}{2}$

Multiply $- 1$ by $- 1$ .

$x = \frac{1 - 1 \pm i \sqrt{7}}{2}$

Multiply $- 1 \pm i \sqrt{7}$ by $1$ .

$x = \frac{- 1 \pm i \sqrt{7}}{2}$

Change the $\pm$ to $+$ .

$x = \frac{- 1 + i \sqrt{7}}{2}$

Rewrite $- 1$ as $- 1 (1)$ .

$x = \frac{- 1 \cdot 1 + i \sqrt{7}}{2}$

Factor $- 1$ out of $i \sqrt{7}$ .

$x = \frac{- 1 \cdot 1 - (- i \sqrt{7})}{2}$

Factor $- 1$ out of $- 1 (1) - (- i \sqrt{7})$ .

$x = \frac{- 1 (1 - i \sqrt{7})}{2}$

Move the negative in front of the fraction.

$x = - \frac{1 - i \sqrt{7}}{2}$

Simplify the expression to solve for the $-$ portion of the $\pm$ .

Tap for more steps...

Simplify the numerator.

Tap for more steps...

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot (1 \cdot 2)}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 2}}{2 \cdot 1}$

Multiply $- 4$ by $2$ .

$x = \frac{- 1 \pm \sqrt{1 - 8}}{2 \cdot 1}$

Subtract $8$ from $1$ .

$x = \frac{- 1 \pm \sqrt{- 7}}{2 \cdot 1}$

Rewrite $- 7$ as $- 1 (7)$ .

$x = \frac{- 1 \pm \sqrt{- 1 \cdot 7}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (7)}$ as $\sqrt{- 1} \cdot \sqrt{7}$ .

$x = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{7}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 1 \pm i \sqrt{7}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm i \sqrt{7}}{2}$

Factor $- 1$ out of $- 1 \pm i \sqrt{7}$ .

$x = \frac{- 11 \pm i \sqrt{7}}{2}$

Multiply $- 1$ by $- 1$ .

$x = \frac{1 - 1 \pm i \sqrt{7}}{2}$

Multiply $- 1 \pm i \sqrt{7}$ by $1$ .

$x = \frac{- 1 \pm i \sqrt{7}}{2}$

Change the $\pm$ to $-$ .

$x = \frac{- 1 - i \sqrt{7}}{2}$

Rewrite $- 1$ as $- 1 (1)$ .

$x = \frac{- 1 \cdot 1 - i \sqrt{7}}{2}$

Factor $- 1$ out of $- i \sqrt{7}$ .

$x = \frac{- 1 \cdot 1 - (i \sqrt{7})}{2}$

Factor $- 1$ out of $- 1 (1) - (i \sqrt{7})$ .

$x = \frac{- 1 (1 + i \sqrt{7})}{2}$

Move the negative in front of the fraction.

$x = - \frac{1 + i \sqrt{7}}{2}$

The final answer is the combination of both solutions.

$x = - \frac{1 - i \sqrt{7}}{2}, - \frac{1 + i \sqrt{7}}{2}$

The solution is the result of $x - 1 = 0$ and $x^{2} + x + 2 = 0$ .

$x = 1, - \frac{1 - i \sqrt{7}}{2}, - \frac{1 + i \sqrt{7}}{2}$

The values which make the derivative equal to $0$ are $1$ .

$1$

After finding the point that makes the derivative $f' (x) = 4 x^{3} + 4 x - 8$ equal to $0$ or undefined, the interval to check where $f (x) = x^{4} + 2 x^{2} - 8 x$ is increasing and where it is decreasing is $(- \infty, 1) \cup (1, \infty)$ .

$(- \infty, 1) \cup (1, \infty)$

Substitute a value from the interval $(- \infty, 1)$ into the derivative to determine if the function is increasing or decreasing.

Tap for more steps...

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

$f' (0) = 4 {(0)}^{3} + 4 (0) - 8$

Simplify the result.

Tap for more steps...

Simplify each term.

Tap for more steps...

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

$f' (0) = 4 \cdot 0 + 4 (0) - 8$

Multiply $4$ by $0$ .

$f' (0) = 0 + 4 (0) - 8$

Multiply $4$ by $0$ .

$f' (0) = 0 + 0 - 8$

Simplify by adding zeros.

Tap for more steps...

Add $0$ and $0$ .

$f' (0) = 0 - 8$

Subtract $8$ from $0$ .

$f' (0) = - 8$

The final answer is $- 8$ .

$- 8$

At $x = 0$ the derivative is $- 8$ . Since this is negative, the function is decreasing on $(- \infty, 1)$ .

Decreasing on $(- \infty, 1)$ since $f' (x) < 0$

Substitute a value from the interval $(1, \infty)$ into the derivative to determine if the function is increasing or decreasing.

Tap for more steps...

Replace the variable $x$ with $2$ in the expression.

$f' (2) = 4 {(2)}^{3} + 4 (2) - 8$

Simplify the result.

Tap for more steps...

Simplify each term.

Tap for more steps...

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

$f' (2) = 4 \cdot 8 + 4 (2) - 8$

Multiply $4$ by $8$ .

$f' (2) = 32 + 4 (2) - 8$

Multiply $4$ by $2$ .

$f' (2) = 32 + 8 - 8$

Simplify by adding and subtracting.

Tap for more steps...

Add $32$ and $8$ .

$f' (2) = 40 - 8$

Subtract $8$ from $40$ .

$f' (2) = 32$

The final answer is $32$ .

$32$

At $x = 2$ the derivative is $32$ . Since this is positive, the function is increasing on $(1, \infty)$ .

Increasing on $(1, \infty)$ since $f' (x) > 0$

List the intervals on which the function is increasing and decreasing.

Increasing on: $(1, \infty)$

Decreasing on: $(- \infty, 1)$

Enter YOUR Problem