Question Find the instantaneous rate of change for $f(x)=\sqrt{2 x+1}$ at $x=4$ .

Studdy Solution

STEP 1

Assumptions
1. The function given is $f(x) = \sqrt{2x + 1}$ .
2. We are interested in the instantaneous rate of change at $x = 4$ .
3. The instantaneous rate of change of a function at a point is given by the derivative of the function evaluated at that point.

STEP 2

To find the instantaneous rate of change of the function at $x = 4$ , we need to find the derivative of the function $f(x)$ with respect to $x$ .
$f'(x) = \frac{d}{dx}(\sqrt{2x + 1})$

STEP 3

To differentiate $\sqrt{2x + 1}$ with respect to $x$ , we will use the chain rule. The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.
$f'(x) = \frac{d}{dx}(\sqrt{2x + 1}) = \frac{1}{2\sqrt{2x + 1}} \cdot \frac{d}{dx}(2x + 1)$

STEP 4

Now, differentiate the inner function $2x + 1$ with respect to $x$ .
$\frac{d}{dx}(2x + 1) = 2$

STEP 5

Substitute the derivative of the inner function back into the expression obtained in STEP_3.
$f'(x) = \frac{1}{2\sqrt{2x + 1}} \cdot 2$

STEP 6

Simplify the expression for the derivative.
$f'(x) = \frac{2}{2\sqrt{2x + 1}}$

STEP 7

Further simplify the expression by canceling the common factor of 2 in the numerator and denominator.
$f'(x) = \frac{1}{\sqrt{2x + 1}}$

STEP 8

Now, evaluate the derivative at $x = 4$ to find the instantaneous rate of change at that point.
$f'(4) = \frac{1}{\sqrt{2(4) + 1}}$

STEP 9

Calculate the value inside the square root.
$f'(4) = \frac{1}{\sqrt{8 + 1}}$

STEP 10

Simplify the square root.
$f'(4) = \frac{1}{\sqrt{9}}$

STEP 11

Since the square root of 9 is 3, we can further simplify the expression.
$f'(4) = \frac{1}{3}$
The instantaneous rate of change for the function $f(x) = \sqrt{2x + 1}$ at $x = 4$ is $\frac{1}{3}$ .
The correct answer is e. $\frac{1}{3}$ .

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