Calculus Examples

$f (x) = x^{4} - 4 x^{3} + 10$

Step 1

Find the second derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

Differentiate.

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Step 1.1.1.1

By the Sum Rule, the derivative of $x^{4} - 4 x^{3} + 10$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 4 x^{3}] + \frac{d}{d x} [10]$ .

$\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 4 x^{3}] + \frac{d}{d x} [10]$

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 $n = 4$ .

$4 x^{3} + \frac{d}{d x} [- 4 x^{3}] + \frac{d}{d x} [10]$

Step 1.1.2

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

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Step 1.1.2.1

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

$4 x^{3} - 4 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [10]$

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 $n = 3$ .

$4 x^{3} - 4 (3 x^{2}) + \frac{d}{d x} [10]$

Step 1.1.2.3

Multiply $3$ by $- 4$ .

$4 x^{3} - 12 x^{2} + \frac{d}{d x} [10]$

Step 1.1.3

Differentiate using the Constant Rule.

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Step 1.1.3.1

Since $10$ is constant with respect to $x$ , the derivative of $10$ with respect to $x$ is $0$ .

$4 x^{3} - 12 x^{2} + 0$

Step 1.1.3.2

Add $4 x^{3} - 12 x^{2}$ and $0$ .

$f' (x) = 4 x^{3} - 12 x^{2}$

Step 1.2

Find the second derivative.

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Step 1.2.1

By the Sum Rule, the derivative of $4 x^{3} - 12 x^{2}$ with respect to $x$ is $\frac{d}{d x} [4 x^{3}] + \frac{d}{d x} [- 12 x^{2}]$ .

$\frac{d}{d x} [4 x^{3}] + \frac{d}{d x} [- 12 x^{2}]$

Step 1.2.2

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

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Step 1.2.2.1

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

$4 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 12 x^{2}]$

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 $n = 3$ .

$4 (3 x^{2}) + \frac{d}{d x} [- 12 x^{2}]$

Step 1.2.2.3

Multiply $3$ by $4$ .

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

Step 1.2.3

Evaluate $\frac{d}{d x} [- 12 x^{2}]$ .

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Step 1.2.3.1

Since $- 12$ is constant with respect to $x$ , the derivative of $- 12 x^{2}$ with respect to $x$ is $- 12 \frac{d}{d x} [x^{2}]$ .

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

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 $n = 2$ .

$12 x^{2} - 12 (2 x)$

Step 1.2.3.3

Multiply $2$ by $- 12$ .

$f'' (x) = 12 x^{2} - 24 x$

Step 1.3

The second derivative of $f (x)$ with respect to $x$ is $12 x^{2} - 24 x$ .

$12 x^{2} - 24 x$

Step 2

Set the second derivative equal to $0$ then solve the equation $12 x^{2} - 24 x = 0$ .

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Step 2.1

Set the second derivative equal to $0$ .

$12 x^{2} - 24 x = 0$

Step 2.2

Factor $12 x$ out of $12 x^{2} - 24 x$ .

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Step 2.2.1

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

$12 x (x) - 24 x = 0$

Step 2.2.2

Factor $12 x$ out of $- 24 x$ .

$12 x (x) + 12 x (- 2) = 0$

Step 2.2.3

Factor $12 x$ out of $12 x (x) + 12 x (- 2)$ .

$12 x (x - 2) = 0$

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 $0$ .

$x = 0$

$x - 2 = 0$

Step 2.4

Set $x$ equal to $0$ .

$x = 0$

Step 2.5

Set $x - 2$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Step 2.5.2

Add $2$ to both sides of the equation.

$x = 2$

Step 2.6

The final solution is all the values that make $12 x (x - 2) = 0$ true.

$x = 0, 2$

Step 3

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

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Step 3.1

Substitute $0$ in $f (x) = x^{4} - 4 x^{3} + 10$ to find the value of $y$ .

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Step 3.1.1

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

$f (0) = {(0)}^{4} - 4 {(0)}^{3} + 10$

Step 3.1.2

Simplify the result.

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Step 3.1.2.1

Simplify each term.

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Step 3.1.2.1.1

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

$f (0) = 0 - 4 {(0)}^{3} + 10$

Step 3.1.2.1.2

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

$f (0) = 0 - 4 \cdot 0 + 10$

Step 3.1.2.1.3

Multiply $- 4$ by $0$ .

$f (0) = 0 + 0 + 10$

Step 3.1.2.2

Simplify by adding numbers.

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Step 3.1.2.2.1

Add $0$ and $0$ .

$f (0) = 0 + 10$

Step 3.1.2.2.2

Add $0$ and $10$ .

$f (0) = 10$

Step 3.1.2.3

The final answer is $10$ .

$10$

Step 3.2

The point found by substituting $0$ in $f (x) = x^{4} - 4 x^{3} + 10$ is $(0, 10)$ . This point can be an inflection point.

$(0, 10)$

Step 3.3

Substitute $2$ in $f (x) = x^{4} - 4 x^{3} + 10$ to find the value of $y$ .

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Step 3.3.1

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

$f (2) = {(2)}^{4} - 4 {(2)}^{3} + 10$

Step 3.3.2

Simplify the result.

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Step 3.3.2.1

Simplify each term.

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Step 3.3.2.1.1

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

$f (2) = 16 - 4 {(2)}^{3} + 10$

Step 3.3.2.1.2

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

$f (2) = 16 - 4 \cdot 8 + 10$

Step 3.3.2.1.3

Multiply $- 4$ by $8$ .

$f (2) = 16 - 32 + 10$

Step 3.3.2.2

Simplify by adding and subtracting.

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Step 3.3.2.2.1

Subtract $32$ from $16$ .

$f (2) = - 16 + 10$

Step 3.3.2.2.2

Add $- 16$ and $10$ .

$f (2) = - 6$

Step 3.3.2.3

The final answer is $- 6$ .

$- 6$

Step 3.4

The point found by substituting $2$ in $f (x) = x^{4} - 4 x^{3} + 10$ is $(2, - 6)$ . This point can be an inflection point.

$(2, - 6)$

Step 3.5

Determine the points that could be inflection points.

$(0, 10), (2, - 6)$

Step 4

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

$(- \infty, 0) \cup (0, 2) \cup (2, \infty)$

Step 5

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

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Step 5.1

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

$f'' (- 0.1) = 12 {(- 0.1)}^{2} - 24 \cdot - 0.1$

Step 5.2

Simplify the result.

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Step 5.2.1

Simplify each term.

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Step 5.2.1.1

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

$f'' (- 0.1) = 12 \cdot 0.01 - 24 \cdot - 0.1$

Step 5.2.1.2

Multiply $12$ by $0.01$ .

$f'' (- 0.1) = 0.12 - 24 \cdot - 0.1$

Step 5.2.1.3

Multiply $- 24$ by $- 0.1$ .

$f'' (- 0.1) = 0.12 + 2.4$

Step 5.2.2

Add $0.12$ and $2.4$ .

$f'' (- 0.1) = 2.52$

Step 5.2.3

The final answer is $2.52$ .

$2.52$

Step 5.3

At $- 0.1$ , the second derivative is $2.52$ . Since this is positive, the second derivative is increasing on the interval $(- \infty, 0)$ .

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

Step 6

Substitute a value from the interval $(0, 2)$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

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

$f'' (1) = 12 {(1)}^{2} - 24 \cdot 1$

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify each term.

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Step 6.2.1.1

One to any power is one.

$f'' (1) = 12 \cdot 1 - 24 \cdot 1$

Step 6.2.1.2

Multiply $12$ by $1$ .

$f'' (1) = 12 - 24 \cdot 1$

Step 6.2.1.3

Multiply $- 24$ by $1$ .

$f'' (1) = 12 - 24$

Step 6.2.2

Subtract $24$ from $12$ .

$f'' (1) = - 12$

Step 6.2.3

The final answer is $- 12$ .

$- 12$

Step 6.3

At $1$ , the second derivative is $- 12$ . Since this is negative, the second derivative is decreasing on the interval $(0, 2)$

Decreasing on $(0, 2)$ since $f'' (x) < 0$

Step 7

Substitute a value from the interval $(2, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $2.1$ in the expression.

$f'' (2.1) = 12 {(2.1)}^{2} - 24 \cdot 2.1$

Step 7.2

Simplify the result.

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Step 7.2.1

Simplify each term.

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Step 7.2.1.1

Raise $2.1$ to the power of $2$ .

$f'' (2.1) = 12 \cdot 4.41 - 24 \cdot 2.1$

Step 7.2.1.2

Multiply $12$ by $4.41$ .

$f'' (2.1) = 52.92 - 24 \cdot 2.1$

Step 7.2.1.3

Multiply $- 24$ by $2.1$ .

$f'' (2.1) = 52.92 - 50.4$

Step 7.2.2

Subtract $50.4$ from $52.92$ .

$f'' (2.1) = 2.52$

Step 7.2.3

The final answer is $2.52$ .

$2.52$

Step 7.3

At $2.1$ , the second derivative is $2.52$ . Since this is positive, the second derivative is increasing on the interval $(2, \infty)$ .

Increasing on $(2, \infty)$ 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 $(0, 10), (2, - 6)$ .

$(0, 10), (2, - 6)$

Step 9