Calculus Examples

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

Find the second derivative.

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Find the first derivative.

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Differentiate.

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

$f' (x) = \frac{d}{d x} (x^{4}) + \frac{d}{d x} (- x^{3})$

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

$f' (x) = 4 x^{3} + \frac{d}{d x} (- x^{3})$

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

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Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{3}$ with respect to $x$ is $- \frac{d}{d x} [x^{3}]$ .

$f' (x) = 4 x^{3} - \frac{d}{d x} x^{3}$

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

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

Multiply $3$ by $- 1$ .

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

Find the second derivative.

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

$f'' (x) = \frac{d}{d x} (4 x^{3}) + \frac{d}{d x} (- 3 x^{2})$

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

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

$f'' (x) = 4 \frac{d}{d x} (x^{3}) + \frac{d}{d x} (- 3 x^{2})$

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

$f'' (x) = 4 (3 x^{2}) + \frac{d}{d x} (- 3 x^{2})$

Multiply $3$ by $4$ .

$f'' (x) = 12 x^{2} + \frac{d}{d x} (- 3 x^{2})$

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

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

$f'' (x) = 12 x^{2} - 3 \frac{d}{d x} x^{2}$

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

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

Multiply $2$ by $- 3$ .

$f'' (x) = 12 x^{2} - 6 x$

The second derivative of $f (x)$ with respect to $x$ is $12 x^{2} - 6 x$ .

$12 x^{2} - 6 x$

Set the second derivative equal to $0$ then solve the equation $12 x^{2} - 6 x = 0$ .

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Set the second derivative equal to $0$ .

$12 x^{2} - 6 x = 0$

Factor $6 x$ out of $12 x^{2} - 6 x$ .

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Factor $6 x$ out of $12 x^{2}$ .

$6 x (2 x) - 6 x = 0$

Factor $6 x$ out of $- 6 x$ .

$6 x (2 x) + 6 x (- 1) = 0$

Factor $6 x$ out of $6 x (2 x) + 6 x (- 1)$ .

$6 x (2 x - 1) = 0$

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

$x = 0$

$2 x - 1 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $2 x - 1$ equal to $0$ and solve for $x$ .

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Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Solve $2 x - 1 = 0$ for $x$ .

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Add $1$ to both sides of the equation.

$2 x = 1$

Divide each term in $2 x = 1$ by $2$ and simplify.

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Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Simplify the left side.

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Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Divide $x$ by $1$ .

$x = \frac{1}{2}$

The final solution is all the values that make $6 x (2 x - 1) = 0$ true.

$x = 0, \frac{1}{2}$

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

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

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

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

Simplify the result.

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Simplify each term.

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

$f (0) = 0 - {(0)}^{3}$

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

$f (0) = 0 - 0$

Multiply $- 1$ by $0$ .

$f (0) = 0 + 0$

Add $0$ and $0$ .

$f (0) = 0$

The final answer is $0$ .

$0$

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

$(0, 0)$

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

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

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

Simplify the result.

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Simplify each term.

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

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

One to any power is one.

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

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

$f (\frac{1}{2}) = \frac{1}{16} - {(\frac{1}{2})}^{3}$

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

$f (\frac{1}{2}) = \frac{1}{16} - \frac{1^{3}}{2^{3}}$

One to any power is one.

$f (\frac{1}{2}) = \frac{1}{16} - \frac{1}{2^{3}}$

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

$f (\frac{1}{2}) = \frac{1}{16} - \frac{1}{8}$

To write $- \frac{1}{8}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f (\frac{1}{2}) = \frac{1}{16} - \frac{1}{8} \cdot \frac{2}{2}$

Write each expression with a common denominator of $16$ , by multiplying each by an appropriate factor of $1$ .

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Multiply $\frac{1}{8}$ by $\frac{2}{2}$ .

$f (\frac{1}{2}) = \frac{1}{16} - \frac{2}{8 \cdot 2}$

Multiply $8$ by $2$ .

$f (\frac{1}{2}) = \frac{1}{16} - \frac{2}{16}$

Combine the numerators over the common denominator.

$f (\frac{1}{2}) = \frac{1 - 2}{16}$

Subtract $2$ from $1$ .

$f (\frac{1}{2}) = \frac{- 1}{16}$

Move the negative in front of the fraction.

$f (\frac{1}{2}) = - \frac{1}{16}$

The final answer is $- \frac{1}{16}$ .

$- \frac{1}{16}$

The point found by substituting $\frac{1}{2}$ in $f (x) = x^{4} - x^{3}$ is $(\frac{1}{2}, - \frac{1}{16})$ . This point can be an inflection point.

$(\frac{1}{2}, - \frac{1}{16})$

Determine the points that could be inflection points.

$(0, 0), (\frac{1}{2}, - \frac{1}{16})$

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

$(- \infty, 0) \cup (0, \frac{1}{2}) \cup (\frac{1}{2}, \infty)$

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

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

$f'' (- 0.1) = 12 {(- 0.1)}^{2} - 6 \cdot - 0.1$

Simplify the result.

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Simplify each term.

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

$f'' (- 0.1) = 12 \cdot 0.01 - 6 \cdot - 0.1$

Multiply $12$ by $0.01$ .

$f'' (- 0.1) = 0.12 - 6 \cdot - 0.1$

Multiply $- 6$ by $- 0.1$ .

$f'' (- 0.1) = 0.12 + 0.6$

Add $0.12$ and $0.6$ .

$f'' (- 0.1) = 0.72$

The final answer is $0.72$ .

$0.72$

At $- 0.1$ , the second derivative is $0.72$ . Since this is positive, the second derivative is increasing on the interval $(- \infty, 0)$ .

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

Substitute a value from the interval $(0, \frac{1}{2})$ into the second derivative to determine if it is increasing or decreasing.

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Replace the variable $x$ with $0.25$ in the expression.

$f'' (0.25) = 12 {(0.25)}^{2} - 6 \cdot 0.25$

Simplify the result.

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Simplify each term.

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Raise $0.25$ to the power of $2$ .

$f'' (0.25) = 12 \cdot 0.0625 - 6 \cdot 0.25$

Multiply $12$ by $0.0625$ .

$f'' (0.25) = 0.75 - 6 \cdot 0.25$

Multiply $- 6$ by $0.25$ .

$f'' (0.25) = 0.75 - 1.5$

Subtract $1.5$ from $0.75$ .

$f'' (0.25) = - 0.75$

The final answer is $- 0.75$ .

$- 0.75$

At $0.25$ , the second derivative is $- 0.75$ . Since this is negative, the second derivative is decreasing on the interval $(0, \frac{1}{2})$

Decreasing on $(0, \frac{1}{2})$ since $f'' (x) < 0$

Substitute a value from the interval $(\frac{1}{2}, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Replace the variable $x$ with $0.6$ in the expression.

$f'' (0.6) = 12 {(0.6)}^{2} - 6 \cdot 0.6$

Simplify the result.

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Simplify each term.

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Raise $0.6$ to the power of $2$ .

$f'' (0.6) = 12 \cdot 0.36 - 6 \cdot 0.6$

Multiply $12$ by $0.36$ .

$f'' (0.6) = 4.32 - 6 \cdot 0.6$

Multiply $- 6$ by $0.6$ .

$f'' (0.6) = 4.32 - 3.6$

Subtract $3.6$ from $4.32$ .

$f'' (0.6) = 0.72$

The final answer is $0.72$ .

$0.72$

At $0.6$ , the second derivative is $0.72$ . Since this is positive, the second derivative is increasing on the interval $(\frac{1}{2}, \infty)$ .

Increasing on $(\frac{1}{2}, \infty)$ since $f'' (x) > 0$

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 $(0, 0), (\frac{1}{2}, - \frac{1}{16})$ .

$(0, 0), (\frac{1}{2}, - \frac{1}{16})$