Calculus Examples

$f (x) = \frac{1}{4} x^{4} + x^{3} + \frac{3}{2} x^{2}$

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

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

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

Step 1.1.2

Evaluate $\frac{d}{d x} [\frac{1}{4} x^{4}]$ .

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Step 1.1.2.1

Since $\frac{1}{4}$ is constant with respect to $x$ , the derivative of $\frac{1}{4} x^{4}$ with respect to $x$ is $\frac{1}{4} \frac{d}{d x} [x^{4}]$ .

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

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 $n = 4$ .

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

Step 1.1.2.3

Combine $4$ and $\frac{1}{4}$ .

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

Step 1.1.2.4

Combine $\frac{4}{4}$ and $x^{3}$ .

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

Step 1.1.2.5

Cancel the common factor of $4$ .

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Step 1.1.2.5.1

Cancel the common factor.

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

Step 1.1.2.5.2

Divide $x^{3}$ by $1$ .

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

Step 1.1.3

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

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

Step 1.1.4

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

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Step 1.1.4.1

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

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

Step 1.1.4.2

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

$x^{3} + 3 x^{2} + \frac{3}{2} (2 x)$

Step 1.1.4.3

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

$x^{3} + 3 x^{2} + \frac{2 \cdot 3}{2} x$

Step 1.1.4.4

Multiply $2$ by $3$ .

$x^{3} + 3 x^{2} + \frac{6}{2} x$

Step 1.1.4.5

Combine $\frac{6}{2}$ and $x$ .

$x^{3} + 3 x^{2} + \frac{6 x}{2}$

Step 1.1.4.6

Cancel the common factor of $6$ and $2$ .

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Step 1.1.4.6.1

Factor $2$ out of $6 x$ .

$x^{3} + 3 x^{2} + \frac{2 (3 x)}{2}$

Step 1.1.4.6.2

Cancel the common factors.

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Step 1.1.4.6.2.1

Factor $2$ out of $2$ .

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

Step 1.1.4.6.2.2

Cancel the common factor.

$x^{3} + 3 x^{2} + \frac{2 (3 x)}{2 \cdot 1}$

Step 1.1.4.6.2.3

Rewrite the expression.

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

Step 1.1.4.6.2.4

Divide $3 x$ by $1$ .

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

Step 1.2

Find the second derivative.

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Step 1.2.1

Differentiate.

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Step 1.2.1.1

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

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

Step 1.2.1.2

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

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

Step 1.2.2

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

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Step 1.2.2.1

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

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

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 $n = 2$ .

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

Step 1.2.2.3

Multiply $2$ by $3$ .

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

Step 1.2.3

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

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Step 1.2.3.1

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

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

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 $n = 1$ .

$3 x^{2} + 6 x + 3 \cdot 1$

Step 1.2.3.3

Multiply $3$ by $1$ .

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

Step 1.3

The second derivative of $f (x)$ with respect to $x$ is $3 x^{2} + 6 x + 3$ .

$3 x^{2} + 6 x + 3$

Step 2

Set the second derivative equal to $0$ then solve the equation $3 x^{2} + 6 x + 3 = 0$ .

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Step 2.1

Set the second derivative equal to $0$ .

$3 x^{2} + 6 x + 3 = 0$

Step 2.2

Factor the left side of the equation.

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Step 2.2.1

Factor $3$ out of $3 x^{2} + 6 x + 3$ .

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Step 2.2.1.1

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

$3 (x^{2}) + 6 x + 3 = 0$

Step 2.2.1.2

Factor $3$ out of $6 x$ .

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

Step 2.2.1.3

Factor $3$ out of $3$ .

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

Step 2.2.1.4

Factor $3$ out of $3 (x^{2}) + 3 (2 x)$ .

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

Step 2.2.1.5

Factor $3$ out of $3 (x^{2} + 2 x) + 3 (1)$ .

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

Step 2.2.2

Factor using the perfect square rule.

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Step 2.2.2.1

Rewrite $1$ as $1^{2}$ .

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

Step 2.2.2.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot x \cdot 1$

Step 2.2.2.3

Rewrite the polynomial.

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

Step 2.2.2.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 1$ .

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

Step 2.3

Divide each term in $3 {(x + 1)}^{2} = 0$ by $3$ and simplify.

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Step 2.3.1

Divide each term in $3 {(x + 1)}^{2} = 0$ by $3$ .

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

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of $3$ .

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Step 2.3.2.1.1

Cancel the common factor.

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

Step 2.3.2.1.2

Divide ${(x + 1)}^{2}$ by $1$ .

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

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Divide $0$ by $3$ .

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

Step 2.4

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

$x + 1 = 0$

Step 2.5

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 3

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

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Step 3.1

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

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Step 3.1.1

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

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

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

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

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

Step 3.1.2.1.2

Multiply $\frac{1}{4}$ by $1$ .

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

Step 3.1.2.1.3

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

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

Step 3.1.2.1.4

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

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

Step 3.1.2.1.5

Multiply $\frac{3}{2}$ by $1$ .

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

Step 3.1.2.2

Find the common denominator.

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Step 3.1.2.2.1

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

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

Step 3.1.2.2.2

Multiply $\frac{- 1}{1}$ by $\frac{4}{4}$ .

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

Step 3.1.2.2.3

Multiply $\frac{- 1}{1}$ by $\frac{4}{4}$ .

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

Step 3.1.2.2.4

Multiply $\frac{3}{2}$ by $\frac{2}{2}$ .

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

Step 3.1.2.2.5

Multiply $\frac{3}{2}$ by $\frac{2}{2}$ .

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

Step 3.1.2.2.6

Multiply $2$ by $2$ .

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

Step 3.1.2.3

Combine the numerators over the common denominator.

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

Step 3.1.2.4

Simplify each term.

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Step 3.1.2.4.1

Multiply $- 1$ by $4$ .

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

Step 3.1.2.4.2

Multiply $3$ by $2$ .

$f (- 1) = \frac{1 - 4 + 6}{4}$

Step 3.1.2.5

Simplify by adding and subtracting.

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Step 3.1.2.5.1

Subtract $4$ from $1$ .

$f (- 1) = \frac{- 3 + 6}{4}$

Step 3.1.2.5.2

Add $- 3$ and $6$ .

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

Step 3.1.2.6

The final answer is $\frac{3}{4}$ .

$\frac{3}{4}$

Step 3.2

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

$(- 1, \frac{3}{4})$

Step 4

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

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

Step 5

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

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Step 5.1

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

$f'' (- 1.1) = 3 {(- 1.1)}^{2} + 6 (- 1.1) + 3$

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

$f'' (- 1.1) = 3 \cdot 1.21 + 6 (- 1.1) + 3$

Step 5.2.1.2

Multiply $3$ by $1.21$ .

$f'' (- 1.1) = 3.63 + 6 (- 1.1) + 3$

Step 5.2.1.3

Multiply $6$ by $- 1.1$ .

$f'' (- 1.1) = 3.63 - 6.6 + 3$

Step 5.2.2

Simplify by adding and subtracting.

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Step 5.2.2.1

Subtract $6.6$ from $3.63$ .

$f'' (- 1.1) = - 2.97 + 3$

Step 5.2.2.2

Add $- 2.97$ and $3$ .

$f'' (- 1.1) = 0.03$

Step 5.2.3

The final answer is $0.03$ .

$0.03$

Step 5.3

At $- 1.1$ , the second derivative is $0.03$ . 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 $(- 1, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $- 0.9$ in the expression.

$f'' (- 0.9) = 3 {(- 0.9)}^{2} + 6 (- 0.9) + 3$

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

Raise $- 0.9$ to the power of $2$ .

$f'' (- 0.9) = 3 \cdot 0.81 + 6 (- 0.9) + 3$

Step 6.2.1.2

Multiply $3$ by $0.81$ .

$f'' (- 0.9) = 2.43 + 6 (- 0.9) + 3$

Step 6.2.1.3

Multiply $6$ by $- 0.9$ .

$f'' (- 0.9) = 2.43 - 5.4 + 3$

Step 6.2.2

Simplify by adding and subtracting.

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Step 6.2.2.1

Subtract $5.4$ from $2.43$ .

$f'' (- 0.9) = - 2.97 + 3$

Step 6.2.2.2

Add $- 2.97$ and $3$ .

$f'' (- 0.9) = 0.03$

Step 6.2.3

The final answer is $0.03$ .

$0.03$

Step 6.3

At $- 0.9$ , the second derivative is $0.03$ . Since this is positive, the second derivative is increasing on the interval $(- 1, \infty)$ .

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

Step 7

An inflection point is a point on a curve at which the concavity changes sign from plus to minus or from minus to plus. There are no points on the graph that satisfy these requirements.

No Inflection Points