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6. \( g(x)=2 \sqrt{x}\left(\frac{1}{5} x^{2}+\frac{1}{3} x+3\right)+C \) correct Explanation: After division \[ g^{\prime}(x)=4 x^{3 / 2}+x^{1 / 2}+3 x^{-1 / 2}, \] so we can now find an antiderivative of each term separately. But \[ \frac{d}{d x}\left(\frac{a x^{r}}{r}\right)=a x^{r-1} \] for all \( a \) and all \( r \neq 0 \). Thus \[ \begin{aligned} \frac{8}{5} x^{5 / 2}+ & \frac{2}{3} x^{3 / 2}+6 x^{1 / 2} \\ & =2 \sqrt{x}\left(\frac{4}{5} x^{2}+\frac{1}{3} x+3\right) \end{aligned} \] - wolesensky - (2022KE) 8 because \( t=0 \) denotes the moment when the brakes are applied, i.e.,

          6. \( g(x)=2 \sqrt{x}\left(\frac{1}{5} x^{2}+\frac{1}{3} x+3\right)+C \) correct

Explanation:
After division
\[
g^{\prime}(x)=4 x^{3 / 2}+x^{1 / 2}+3 x^{-1 / 2},
\]
so we can now find an antiderivative of each term separately. But
\[
\frac{d}{d x}\left(\frac{a x^{r}}{r}\right)=a x^{r-1}
\]
for all \( a \) and all \( r \neq 0 \). Thus
\[
\begin{aligned}
\frac{8}{5} x^{5 / 2}+ & \frac{2}{3} x^{3 / 2}+6 x^{1 / 2} \\
& =2 \sqrt{x}\left(\frac{4}{5} x^{2}+\frac{1}{3} x+3\right)
\end{aligned}
\]
- wolesensky - (2022KE)
8
because \( t=0 \) denotes the moment when the brakes are applied, i.e.,

6. g(x)=2 √(x)((1)/(5) x^2+(1)/(3) x+3)+C correct

Explanation:
After division

g^'(x)=4 x^3 / 2+x^1 / 2+3 x^-1 / 2,

so we can now find an antiderivative of each term separately. But

(d)/(d x)((a x^r)/(r))=a x^r-1

for all a and all r ≠ 0. Thus

(8)/(5) x^5 / 2+ (2)/(3) x^3 / 2+6 x^1 / 2
=2 √(x)((4)/(5) x^2+(1)/(3) x+3)

- wolesensky - (2022KE)
8
because t=0 denotes the moment when the brakes are applied, i.e.,

Added by Margaret P.

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