Question

1. -/1 points if f(x) = \frac{x^{2}}{4 + x}, find f(2). f(2) =

Explanation:

Step1: Apply quotient - rule for first derivative

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = x^{2}$, $u^\prime=2x$, $v = 4 + x$, and $v^\prime = 1$. So, $f^\prime(x)=\frac{2x(4 + x)-x^{2}(1)}{(4 + x)^{2}}=\frac{8x+2x^{2}-x^{2}}{(4 + x)^{2}}=\frac{x^{2}+8x}{(4 + x)^{2}}$.

Step2: Apply quotient - rule for second derivative

Let $u=x^{2}+8x$, $u^\prime = 2x + 8$, $v=(4 + x)^{2}$, and $v^\prime=2(4 + x)$. Then $f^{\prime\prime}(x)=\frac{(2x + 8)(4 + x)^{2}-(x^{2}+8x)\times2(4 + x)}{(4 + x)^{4}}$. Factor out $(4 + x)$ from the numerator: $f^{\prime\prime}(x)=\frac{(4 + x)[(2x + 8)(4 + x)-2(x^{2}+8x)]}{(4 + x)^{4}}=\frac{(2x + 8)(4 + x)-2(x^{2}+8x)}{(4 + x)^{3}}$. Expand the numerator: $(2x + 8)(4 + x)=8x+2x^{2}+32 + 8x=2x^{2}+16x + 32$, and $2(x^{2}+8x)=2x^{2}+16x$. So, $f^{\prime\prime}(x)=\frac{2x^{2}+16x + 32-(2x^{2}+16x)}{(4 + x)^{3}}=\frac{32}{(4 + x)^{3}}$.

Step3: Evaluate $f^{\prime\prime}(2)$

Substitute $x = 2$ into $f^{\prime\prime}(x)$: $f^{\prime\prime}(2)=\frac{32}{(4 + 2)^{3}}=\frac{32}{216}=\frac{4}{27}$.

Answer:

$\frac{4}{27}$