Calculus Examples

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

Find the inflection points.

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Find the second derivative.

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Find the first derivative.

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

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

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

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

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

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

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

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

Multiply $3$ by $- 4$ .

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

Find the second derivative.

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

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

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

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

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

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

Multiply $3$ by $4$ .

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

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

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

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

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

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

Multiply $2$ by $- 12$ .

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

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

$12 x^{2} - 24 x$

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

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

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

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

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

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

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

$12 x (x - 2) = 0$

Divide each term by $12$ and simplify.

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Divide each term in $12 x = 0$ by $12$ .

$\frac{12 x}{12} = \frac{0}{12}$

Reduce the expression by cancelling the common factors.

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Cancel the common factor.

$\frac{12 x}{12} = \frac{0}{12}$

Divide $x$ by $1$ .

$x = \frac{0}{12}$

Divide $0$ by $12$ .

$x = 0$

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

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Set the factor equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

The solution is the result of $12 x = 0$ and $x - 2 = 0$ .

$x = 0, 2$

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

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

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

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

Simplify the result.

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Simplify each term.

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Remove parentheses.

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

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

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

Remove parentheses.

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

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

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

Multiply $- 4$ 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 $f (x) = x^{4} - 4 x^{3}$ is $(0, 0)$ . This point can be an inflection point.

$(0, 0)$

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

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

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

Simplify the result.

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Simplify each term.

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Remove parentheses.

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

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

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

Remove parentheses.

$f (2) = 16 - 4 \cdot 2^{3}$

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

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

Multiply $- 4$ by $8$ .

$f (2) = 16 - 32$

Subtract $32$ from $16$ .

$f (2) = - 16$

The final answer is $- 16$ .

$- 16$

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

$(2, - 16)$

Determine the points that could be inflection points.

$(0, 0), (2, - 16)$

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

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

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.

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

Simplify the result.

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Simplify each term.

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

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

Multiply $12$ by $0.01$ .

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

Multiply $- 24$ by $- 0.1$ .

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

Add $0.12$ and $2.4$ .

$f'' (- 0.1) = 2.52$

The final answer is $2.52$ .

$2.52$

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 $f'' (x) \to 0$

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

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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.

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

Simplify the result.

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Simplify each term.

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

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

Multiply $12$ by $0.01$ .

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

Multiply $- 24$ by $- 0.1$ .

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

Add $0.12$ and $2.4$ .

$f'' (- 0.1) = 2.52$

The final answer is $2.52$ .

$2.52$

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 $f'' (x) \to 0$

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

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

Simplify the result.

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Simplify each term.

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Remove parentheses.

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

One to any power is one.

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

Multiply $12$ by $1$ .

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

Multiply $- 24$ by $1$ .

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

Subtract $24$ from $12$ .

$f'' (1) = - 12$

The final answer is $- 12$ .

$- 12$

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 $f'' (x) = < 0$

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

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

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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.

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

Simplify the result.

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Simplify each term.

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

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

Multiply $12$ by $0.01$ .

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

Multiply $- 24$ by $- 0.1$ .

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

Add $0.12$ and $2.4$ .

$f'' (- 0.1) = 2.52$

The final answer is $2.52$ .

$2.52$

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 $f'' (x) \to 0$

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

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

Simplify the result.

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Simplify each term.

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Remove parentheses.

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

One to any power is one.

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

Multiply $12$ by $1$ .

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

Multiply $- 24$ by $1$ .

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

Subtract $24$ from $12$ .

$f'' (1) = - 12$

The final answer is $- 12$ .

$- 12$

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 $f'' (x) = < 0$

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

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

Simplify the result.

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Simplify each term.

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Remove parentheses.

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

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

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

Multiply $12$ by $4.41$ .

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

Multiply $- 24$ by $2.1$ .

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

Subtract $50.4$ from $52.92$ .

$f'' (2.1) = 2.52$

The final answer is $2.52$ .

$2.52$

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 $f'' (x) \to 0$

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 $(0, 0), (2, - 16)$ .

$(0, 0), (2, - 16)$

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

$(- \infty, \infty)$

${x | x \in ℝ}$

Create intervals around the inflection points and the undefined values.

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

Substitute any number from the interval $(- \infty, 0)$ into the second derivative and evaluate to determine the concavity.

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

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

Simplify the result.

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Simplify each term.

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

$f'' (- 2) = 12 \cdot 4 - 24 \cdot - 2$

Multiply $12$ by $4$ .

$f'' (- 2) = 48 - 24 \cdot - 2$

Multiply $- 24$ by $- 2$ .

$f'' (- 2) = 48 + 48$

Add $48$ and $48$ .

$f'' (- 2) = 96$

The final answer is $96$ .

$96$

The graph is concave up on the interval $(- \infty, 0)$ because $f'' (- 2)$ is positive.

Concave up on $(- \infty, 0)$ since $f'' (x)$ is positive

Substitute any number from the interval $(0, 2)$ into the second derivative and evaluate to determine the concavity.

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

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

Simplify the result.

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Simplify each term.

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Remove parentheses.

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

One to any power is one.

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

Multiply $12$ by $1$ .

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

Multiply $- 24$ by $1$ .

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

Subtract $24$ from $12$ .

$f'' (1) = - 12$

The final answer is $- 12$ .

$- 12$

The graph is concave down on the interval $(0, 2)$ because $f'' (1)$ is negative.

Concave down on $(0, 2)$ since $f'' (x)$ is negative

Substitute any number from the interval $(2, \infty)$ into the second derivative and evaluate to determine the concavity.

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Replace the variable $x$ with $4$ in the expression.

$f'' (4) = 12 {(4)}^{2} - 24 \cdot 4$

Simplify the result.

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Simplify each term.

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Remove parentheses.

$f'' (4) = 12 \cdot 4^{2} - 24 \cdot 4$

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

$f'' (4) = 12 \cdot 16 - 24 \cdot 4$

Multiply $12$ by $16$ .

$f'' (4) = 192 - 24 \cdot 4$

Multiply $- 24$ by $4$ .

$f'' (4) = 192 - 96$

Subtract $96$ from $192$ .

$f'' (4) = 96$

The final answer is $96$ .

$96$

The graph is concave up on the interval $(2, \infty)$ because $f'' (4)$ is positive.

Concave up on $(2, \infty)$ since $f'' (x)$ is positive

The graph is concave down when the second derivative is negative and concave up when the second derivative is positive.

Concave up on $(- \infty, 0)$ since $f'' (x)$ is positive

Concave down on $(0, 2)$ since $f'' (x)$ is negative

Concave up on $(2, \infty)$ since $f'' (x)$ is positive

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