Calculus Examples

$f (x) = x^{4} + 2 x^{3} - 8 x + 1$

Step 1

Find the second derivative.

Tap for more steps...

Step 1.1

Find the first derivative.

Tap for more steps...

Step 1.1.1

Differentiate.

Tap for more steps...

Step 1.1.1.1

By the Sum Rule, the derivative of $x^{4} + 2 x^{3} - 8 x + 1$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [2 x^{3}] + \frac{d}{d x} [- 8 x] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [x^{4}] + \frac{d}{d x} [2 x^{3}] + \frac{d}{d x} [- 8 x] + \frac{d}{d x} [1]$

Step 1.1.1.2

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

$4 x^{3} + \frac{d}{d x} [2 x^{3}] + \frac{d}{d x} [- 8 x] + \frac{d}{d x} [1]$

Step 1.1.2

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

Tap for more steps...

Step 1.1.2.1

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

$4 x^{3} + 2 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 8 x] + \frac{d}{d x} [1]$

Step 1.1.2.2

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

$4 x^{3} + 2 (3 x^{2}) + \frac{d}{d x} [- 8 x] + \frac{d}{d x} [1]$

Step 1.1.2.3

Multiply $3$ by $2$ .

$4 x^{3} + 6 x^{2} + \frac{d}{d x} [- 8 x] + \frac{d}{d x} [1]$

Step 1.1.3

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

Tap for more steps...

Step 1.1.3.1

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

$4 x^{3} + 6 x^{2} - 8 \frac{d}{d x} [x] + \frac{d}{d x} [1]$

Step 1.1.3.2

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

$4 x^{3} + 6 x^{2} - 8 \cdot 1 + \frac{d}{d x} [1]$

Step 1.1.3.3

Multiply $- 8$ by $1$ .

$4 x^{3} + 6 x^{2} - 8 + \frac{d}{d x} [1]$

Step 1.1.4

Differentiate using the Constant Rule.

Tap for more steps...

Step 1.1.4.1

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

$4 x^{3} + 6 x^{2} - 8 + 0$

Step 1.1.4.2

Add $4 x^{3} + 6 x^{2} - 8$ and $0$ .

$f' (x) = 4 x^{3} + 6 x^{2} - 8$

Step 1.2

Find the second derivative.

Tap for more steps...

Step 1.2.1

By the Sum Rule, the derivative of $4 x^{3} + 6 x^{2} - 8$ with respect to $x$ is $\frac{d}{d x} [4 x^{3}] + \frac{d}{d x} [6 x^{2}] + \frac{d}{d x} [- 8]$ .

$\frac{d}{d x} [4 x^{3}] + \frac{d}{d x} [6 x^{2}] + \frac{d}{d x} [- 8]$

Step 1.2.2

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

Tap for more steps...

Step 1.2.2.1

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

$4 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [6 x^{2}] + \frac{d}{d x} [- 8]$

Step 1.2.2.2

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

$4 (3 x^{2}) + \frac{d}{d x} [6 x^{2}] + \frac{d}{d x} [- 8]$

Step 1.2.2.3

Multiply $3$ by $4$ .

$12 x^{2} + \frac{d}{d x} [6 x^{2}] + \frac{d}{d x} [- 8]$

Step 1.2.3

Evaluate $\frac{d}{d x} [6 x^{2}]$ .

Tap for more steps...

Step 1.2.3.1

Since $6$ is constant with respect to $x$ , the derivative of $6 x^{2}$ with respect to $x$ is $6 \frac{d}{d x} [x^{2}]$ .

$12 x^{2} + 6 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 8]$

Step 1.2.3.2

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

$12 x^{2} + 6 (2 x) + \frac{d}{d x} [- 8]$

Step 1.2.3.3

Multiply $2$ by $6$ .

$12 x^{2} + 12 x + \frac{d}{d x} [- 8]$

Step 1.2.4

Differentiate using the Constant Rule.

Tap for more steps...

Step 1.2.4.1

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8$ with respect to $x$ is $0$ .

$12 x^{2} + 12 x + 0$

Step 1.2.4.2

Add $12 x^{2} + 12 x$ and $0$ .

$f'' (x) = 12 x^{2} + 12 x$

Step 1.3

The second derivative of $f (x)$ with respect to $x$ is $12 x^{2} + 12 x$ .

$12 x^{2} + 12 x$

Step 2

Set the second derivative equal to $0$ then solve the equation $12 x^{2} + 12 x = 0$ .

Tap for more steps...

Step 2.1

Set the second derivative equal to $0$ .

$12 x^{2} + 12 x = 0$

Step 2.2

Factor $12 x$ out of $12 x^{2} + 12 x$ .

Tap for more steps...

Step 2.2.1

Factor $12 x$ out of $12 x^{2}$ .

$12 x (x) + 12 x = 0$

Step 2.2.2

Factor $12 x$ out of $12 x$ .

$12 x (x) + 12 x (1) = 0$

Step 2.2.3

Factor $12 x$ out of $12 x (x) + 12 x (1)$ .

$12 x (x + 1) = 0$

Step 2.3

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

$x = 0$

$x + 1 = 0$

Step 2.4

Set $x$ equal to $0$ .

$x = 0$

Step 2.5

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

Tap for more steps...

Step 2.5.1

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

$x + 1 = 0$

Step 2.5.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.6

The final solution is all the values that make $12 x (x + 1) = 0$ true.

$x = 0, - 1$

Step 3

Find the points where the second derivative is $0$ .

Tap for more steps...

Step 3.1

Substitute $0$ in $f (x) = x^{4} + 2 x^{3} - 8 x + 1$ to find the value of $y$ .

Tap for more steps...

Step 3.1.1

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

$f (0) = {(0)}^{4} + 2 {(0)}^{3} - 8 \cdot 0 + 1$

Step 3.1.2

Simplify the result.

Tap for more steps...

Step 3.1.2.1

Simplify each term.

Tap for more steps...

Step 3.1.2.1.1

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

$f (0) = 0 + 2 {(0)}^{3} - 8 \cdot 0 + 1$

Step 3.1.2.1.2

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

$f (0) = 0 + 2 \cdot 0 - 8 \cdot 0 + 1$

Step 3.1.2.1.3

Multiply $2$ by $0$ .

$f (0) = 0 + 0 - 8 \cdot 0 + 1$

Step 3.1.2.1.4

Multiply $- 8$ by $0$ .

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

Step 3.1.2.2

Simplify by adding numbers.

Tap for more steps...

Step 3.1.2.2.1

Add $0$ and $0$ .

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

Step 3.1.2.2.2

Add $0$ and $0$ .

$f (0) = 0 + 1$

Step 3.1.2.2.3

Add $0$ and $1$ .

$f (0) = 1$

Step 3.1.2.3

The final answer is $1$ .

$1$

Step 3.2

The point found by substituting $0$ in $f (x) = x^{4} + 2 x^{3} - 8 x + 1$ is $(0, 1)$ . This point can be an inflection point.

$(0, 1)$

Step 3.3

Substitute $- 1$ in $f (x) = x^{4} + 2 x^{3} - 8 x + 1$ to find the value of $y$ .

Tap for more steps...

Step 3.3.1

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

$f (- 1) = {(- 1)}^{4} + 2 {(- 1)}^{3} - 8 \cdot - 1 + 1$

Step 3.3.2

Simplify the result.

Tap for more steps...

Step 3.3.2.1

Simplify each term.

Tap for more steps...

Step 3.3.2.1.1

Raise $- 1$ to the power of $4$ .

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

Step 3.3.2.1.2

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

$f (- 1) = 1 + 2 \cdot - 1 - 8 \cdot - 1 + 1$

Step 3.3.2.1.3

Multiply $2$ by $- 1$ .

$f (- 1) = 1 - 2 - 8 \cdot - 1 + 1$

Step 3.3.2.1.4

Multiply $- 8$ by $- 1$ .

$f (- 1) = 1 - 2 + 8 + 1$

Step 3.3.2.2

Simplify by adding and subtracting.

Tap for more steps...

Step 3.3.2.2.1

Subtract $2$ from $1$ .

$f (- 1) = - 1 + 8 + 1$

Step 3.3.2.2.2

Add $- 1$ and $8$ .

$f (- 1) = 7 + 1$

Step 3.3.2.2.3

Add $7$ and $1$ .

$f (- 1) = 8$

Step 3.3.2.3

The final answer is $8$ .

$8$

Step 3.4

The point found by substituting $- 1$ in $f (x) = x^{4} + 2 x^{3} - 8 x + 1$ is $(- 1, 8)$ . This point can be an inflection point.

$(- 1, 8)$

Step 3.5

Determine the points that could be inflection points.

$(0, 1), (- 1, 8)$

Step 4

Split $(- \infty, \infty)$ into intervals around the points that could potentially be inflection points.

$(- \infty, - 1) \cup (- 1, 0) \cup (0, \infty)$

Step 5

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

Tap for more steps...

Step 5.1

Replace the variable $x$ with $- 1.1$ in the expression.

$f'' (- 1.1) = 12 {(- 1.1)}^{2} + 12 (- 1.1)$

Step 5.2

Simplify the result.

Tap for more steps...

Step 5.2.1

Simplify each term.

Tap for more steps...

Step 5.2.1.1

Raise $- 1.1$ to the power of $2$ .

$f'' (- 1.1) = 12 \cdot 1.21 + 12 (- 1.1)$

Step 5.2.1.2

Multiply $12$ by $1.21$ .

$f'' (- 1.1) = 14.52 + 12 (- 1.1)$

Step 5.2.1.3

Multiply $12$ by $- 1.1$ .

$f'' (- 1.1) = 14.52 - 13.2$

Step 5.2.2

Subtract $13.2$ from $14.52$ .

$f'' (- 1.1) = 1.32$

Step 5.2.3

The final answer is $1.32$ .

$1.32$

Step 5.3

At $- 1.1$ , the second derivative is $1.32$ . Since this is positive, the second derivative is increasing on the interval $(- \infty, - 1)$ .

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

Step 6

Substitute a value from the interval $(- 1, 0)$ into the second derivative to determine if it is increasing or decreasing.

Tap for more steps...

Step 6.1

Replace the variable $x$ with $- \frac{1}{2}$ in the expression.

$f'' (- \frac{1}{2}) = 12 {(- \frac{1}{2})}^{2} + 12 (- \frac{1}{2})$

Step 6.2

Simplify the result.

Tap for more steps...

Step 6.2.1

Simplify each term.

Tap for more steps...

Step 6.2.1.1

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Tap for more steps...

Step 6.2.1.1.1

Apply the product rule to $- \frac{1}{2}$ .

$f'' (- \frac{1}{2}) = 12 ({(- 1)}^{2} {(\frac{1}{2})}^{2}) + 12 (- \frac{1}{2})$

Step 6.2.1.1.2

Apply the product rule to $\frac{1}{2}$ .

$f'' (- \frac{1}{2}) = 12 ({(- 1)}^{2} (\frac{1^{2}}{2^{2}})) + 12 (- \frac{1}{2})$

Step 6.2.1.2

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

$f'' (- \frac{1}{2}) = 12 (1 (\frac{1^{2}}{2^{2}})) + 12 (- \frac{1}{2})$

Step 6.2.1.3

Multiply $\frac{1^{2}}{2^{2}}$ by $1$ .

$f'' (- \frac{1}{2}) = 12 (\frac{1^{2}}{2^{2}}) + 12 (- \frac{1}{2})$

Step 6.2.1.4

One to any power is one.

$f'' (- \frac{1}{2}) = 12 (\frac{1}{2^{2}}) + 12 (- \frac{1}{2})$

Step 6.2.1.5

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

$f'' (- \frac{1}{2}) = 12 (\frac{1}{4}) + 12 (- \frac{1}{2})$

Step 6.2.1.6

Cancel the common factor of $4$ .

Tap for more steps...

Step 6.2.1.6.1

Factor $4$ out of $12$ .

$f'' (- \frac{1}{2}) = 4 (3) (\frac{1}{4}) + 12 (- \frac{1}{2})$

Step 6.2.1.6.2

Cancel the common factor.

$f'' (- \frac{1}{2}) = 4 \cdot (3 (\frac{1}{4})) + 12 (- \frac{1}{2})$

Step 6.2.1.6.3

Rewrite the expression.

$f'' (- \frac{1}{2}) = 3 + 12 (- \frac{1}{2})$

Step 6.2.1.7

Cancel the common factor of $2$ .

Tap for more steps...

Step 6.2.1.7.1

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$f'' (- \frac{1}{2}) = 3 + 12 (\frac{- 1}{2})$

Step 6.2.1.7.2

Factor $2$ out of $12$ .

$f'' (- \frac{1}{2}) = 3 + 2 (6) (\frac{- 1}{2})$

Step 6.2.1.7.3

Cancel the common factor.

$f'' (- \frac{1}{2}) = 3 + 2 \cdot (6 (\frac{- 1}{2}))$

Step 6.2.1.7.4

Rewrite the expression.

$f'' (- \frac{1}{2}) = 3 + 6 \cdot - 1$

Step 6.2.1.8

Multiply $6$ by $- 1$ .

$f'' (- \frac{1}{2}) = 3 - 6$

Step 6.2.2

Subtract $6$ from $3$ .

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

Step 6.2.3

The final answer is $- 3$ .

$- 3$

Step 6.3

At $- \frac{1}{2}$ , the second derivative is $- 3$ . Since this is negative, the second derivative is decreasing on the interval $(- 1, 0)$

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

Step 7

Substitute a value from the interval $(0, \infty)$ into the second derivative to determine if it is increasing or decreasing.

Tap for more steps...

Step 7.1

Replace the variable $x$ with $0.1$ in the expression.

$f'' (0.1) = 12 {(0.1)}^{2} + 12 (0.1)$

Step 7.2

Simplify the result.

Tap for more steps...

Step 7.2.1

Simplify each term.

Tap for more steps...

Step 7.2.1.1

Raise $0.1$ to the power of $2$ .

$f'' (0.1) = 12 \cdot 0.01 + 12 (0.1)$

Step 7.2.1.2

Multiply $12$ by $0.01$ .

$f'' (0.1) = 0.12 + 12 (0.1)$

Step 7.2.1.3

Multiply $12$ by $0.1$ .

$f'' (0.1) = 0.12 + 1.2$

Step 7.2.2

Add $0.12$ and $1.2$ .

$f'' (0.1) = 1.32$

Step 7.2.3

The final answer is $1.32$ .

$1.32$

Step 7.3

At $0.1$ , the second derivative is $1.32$ . Since this is positive, the second derivative is increasing on the interval $(0, \infty)$ .

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

Step 8

An inflection point is a point on a curve at which the concavity changes sign from plus to minus or from minus to plus. The inflection points in this case are $(- 1, 8), (0, 1)$ .

$(- 1, 8), (0, 1)$

Step 9

Enter YOUR Problem