Question

find f(1) if f(x) = (6x^2 + 8x + 3)/sqrt(x). f(1) = (round to two decimal places as needed.) video example: solving a similar problem a 210_3.4_q8_sum25 find f(4) if f(x) = (4x^2 + 3x + 5)/sqrt(x). u = 4x^2+3x + 5 v = sqrt(x) u = 8x + 3 f(x)=(8x + 3)(sqrt(x))-((1)/(2sqrt(x)))(4x^2+3x + 5)/x

Explanation:

Step1: Rewrite the function

Rewrite $f(x)=\frac{6x^{2}+8x + 3}{\sqrt{x}}$ as $f(x)=6x^{\frac{3}{2}}+8x^{\frac{1}{2}}+3x^{-\frac{1}{2}}$ using the rule $\frac{a + b + c}{d}=\frac{a}{d}+\frac{b}{d}+\frac{c}{d}$ and $\frac{x^{n}}{\sqrt{x}}=x^{n-\frac{1}{2}}$.

Step2: Differentiate term - by - term

Using the power rule $(x^{n})'=nx^{n - 1}$, we have:
$f'(x)=6\times\frac{3}{2}x^{\frac{3}{2}-1}+8\times\frac{1}{2}x^{\frac{1}{2}-1}+3\times(-\frac{1}{2})x^{-\frac{1}{2}-1}$
$f'(x)=9x^{\frac{1}{2}} + 4x^{-\frac{1}{2}}-\frac{3}{2}x^{-\frac{3}{2}}$

Step3: Evaluate $f'(1)$

Substitute $x = 1$ into $f'(x)$:
$f'(1)=9\times1^{\frac{1}{2}}+4\times1^{-\frac{1}{2}}-\frac{3}{2}\times1^{-\frac{3}{2}}$
$f'(1)=9 + 4-\frac{3}{2}$
$f'(1)=\frac{18 + 8-3}{2}=\frac{23}{2}=11.50$

Answer:

$11.50$