Calculus Examples

$y = {(0.3)}^{x}$ , $[- 2, 2]$

Step 1

Find the critical points.

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Step 1.1

Find the first derivative.

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Step 1.1.1

Differentiate using the Exponential Rule which states that $\frac{d}{d x} [a^{x}]$ is $a^{x} \ln (a)$ where $a$ = $0.3$ .

$f' (x) = {0.3}^{x} \ln (0.3)$

Step 1.1.2

The first derivative of $f (x)$ with respect to $x$ is ${0.3}^{x} \ln (0.3)$ .

${0.3}^{x} \ln (0.3)$

Step 1.2

Set the first derivative equal to $0$ then solve the equation ${0.3}^{x} \ln (0.3) = 0$ .

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Step 1.2.1

Set the first derivative equal to $0$ .

${0.3}^{x} \ln (0.3) = 0$

Step 1.2.2

Graph each side of the equation. The solution is the x-value of the point of intersection.

No solution

Step 1.3

Find the values where the derivative is undefined.

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Step 1.3.1

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.

Step 1.4

There are no values of $x$ in the domain of the original problem where the derivative is $0$ or undefined.

No critical points found

Step 2

Evaluate at the included endpoints.

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Step 2.1

Evaluate at $x = - 2$ .

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Step 2.1.1

Substitute $- 2$ for $x$ .

${(0.3)}^{- 2}$

Step 2.1.2

Simplify.

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Step 2.1.2.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{{0.3}^{2}}$

Step 2.1.2.2

Raise $0.3$ to the power of $2$ .

$\frac{1}{0.09}$

Step 2.1.2.3

Divide $1$ by $0.09$ .

$11.\overline{1}$

Step 2.2

Evaluate at $x = 2$ .

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Step 2.2.1

Substitute $2$ for $x$ .

${(0.3)}^{2}$

Step 2.2.2

Raise $0.3$ to the power of $2$ .

$0.09$

Step 2.3

List all of the points.

$(- 2, 11.\overline{1}), (2, 0.09)$

Step 3

Compare the $f (x)$ values found for each value of $x$ in order to determine the absolute maximum and minimum over the given interval. The maximum will occur at the highest $f (x)$ value and the minimum will occur at the lowest $f (x)$ value.

Absolute Maximum: $(- 2, 11.\overline{1})$

Absolute Minimum: $(2, 0.09)$

Step 4