QuestionFind $f(x)$ such that $f(x)=\frac{7}{x}+3 \sqrt{x}$ and $\begin{array}{l} f(1)=13 \\ \text { A. } f(x)=7 \ln |x|+2 x^{3 / 2}+4 \\ B \cdot f(x)=7 \ln |x|+2 x^{3 / 2}+1 \mid \\ \text { C. } f(x)=\frac{7}{x^{2}}+3 x^{3 / 2}+3 \\ \text { D. } f(x)=7 \ln |x|+3 x^{3 / 2}+10 \\ E \cdot f(x)=\frac{7}{x^{2}}+2 x^{3 / 2}+4 \end{array}$

Studdy Solution

STEP 1

1. We are given the function $f(x) = \frac{7}{x} + 3\sqrt{x}$ .
2. We need to find which option satisfies $f(1) = 13$ .

STEP 2

1. Evaluate the given function $f(x)$ at $x = 1$ .
2. Check each option to see if it satisfies the condition $f(1) = 13$ .

STEP 3

Evaluate $f(x) = \frac{7}{x} + 3\sqrt{x}$ at $x = 1$ .
$f(1) = \frac{7}{1} + 3\sqrt{1} = 7 + 3 = 10$

STEP 4

Check each option to see if it satisfies $f(1) = 13$ .
Option A: $f(x) = 7 \ln |x| + 2x^{3/2} + 4$
$f(1) = 7 \ln |1| + 2(1)^{3/2} + 4 = 0 + 2 + 4 = 6$
Option B: $f(x) = 7 \ln |x| + 2x^{3/2} + 1$
$f(1) = 7 \ln |1| + 2(1)^{3/2} + 1 = 0 + 2 + 1 = 3$
Option C: $f(x) = \frac{7}{x^2} + 3x^{3/2} + 3$
$f(1) = \frac{7}{1^2} + 3(1)^{3/2} + 3 = 7 + 3 + 3 = 13$
Option D: $f(x) = 7 \ln |x| + 3x^{3/2} + 10$
$f(1) = 7 \ln |1| + 3(1)^{3/2} + 10 = 0 + 3 + 10 = 13$
Option E: $f(x) = \frac{7}{x^2} + 2x^{3/2} + 4$
$f(1) = \frac{7}{1^2} + 2(1)^{3/2} + 4 = 7 + 2 + 4 = 13$
The options that satisfy $f(1) = 13$ are:
$\boxed{\text{C, D, and E}}$

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