Calculus Examples

$f (x) = x^{4} - 6$

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 $x^{4} - 6$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 6]$ .

$f' (x) = \frac{d}{d x} (x^{4}) + \frac{d}{d x} (- 6)$

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

$f' (x) = 4 x^{3} + \frac{d}{d x} (- 6)$

Since $- 6$ is constant with respect to $x$ , the derivative of $- 6$ with respect to $x$ is $0$ .

$f' (x) = 4 x^{3} + 0$

Add $4 x^{3}$ and $0$ .

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

Find the second derivative.

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

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

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

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

Multiply $3$ by $4$ .

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

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

$12 x^{2}$

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

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Set the second derivative equal to $0$ .

$12 x^{2} = 0$

Divide each term by $12$ and simplify.

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Divide each term in $12 x^{2} = 0$ by $12$ .

$\frac{12 x^{2}}{12} = \frac{0}{12}$

Cancel the common factor of $12$ .

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Cancel the common factor.

$\frac{12 x^{2}}{12} = \frac{0}{12}$

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{0}{12}$

Divide $0$ by $12$ .

$x^{2} = 0$

Take the $square$ root of both sides of the $equation$ to eliminate the exponent on the left side.

$x = \pm \sqrt{0}$

The complete solution is the result of both the positive and negative portions of the solution.

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Simplify the right side of the equation.

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Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Pull terms out from under the radical.

$x = \pm | 0 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $0$ is $0$ .

$x = \pm 0$

$\pm 0$ is equal to $0$ .

$x = 0$

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

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Substitute $0$ in $f (x) = x^{4} - 6$ to find the value of $y$ .

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

$f (0) = {(0)}^{4} - 6$

Simplify the result.

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

$f (0) = 0 - 6$

Subtract $6$ from $0$ .

$f (0) = - 6$

The final answer is $- 6$ .

$- 6$

The point found by substituting $0$ in $f (x) = x^{4} - 6$ is $(0, - 6)$ . This point can be an inflection point.

$(0, - 6)$

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

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

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.

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

Simplify the result.

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

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

Multiply $12$ by $0.01$ .

$f'' (- 0.1) = 0.12$

The final answer is $0.12$ .

$0.12$

At $- 0.1$ , the second derivative is $0.12$ . 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 $(0, \infty)$ 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.

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

Simplify the result.

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

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

Multiply $12$ by $0.01$ .

$f'' (0.1) = 0.12$

The final answer is $0.12$ .

$0.12$

At $0.1$ , the second derivative is $0.12$ . Since this is positive, the second derivative is increasing on the interval $(0, \infty)$ .

Increasing on $(0, \infty)$ 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. There are no points on the graph that satisfy these requirements.

No Inflection Points

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.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

The graph is concave up because the second derivative is positive.

The graph is concave up

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