Question

suppose that
find (f(2)).
(f(x)=\frac{x + 2}{e^{x}})

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = x + 2$ and $v=e^{x}$.

Step2: Find $u'$ and $v'$

Differentiate $u=x + 2$ with respect to $x$, we get $u'=1$. Differentiate $v = e^{x}$ with respect to $x$, we get $v'=e^{x}$.

Step3: Apply the quotient - rule

$f'(x)=\frac{(x + 2)'\cdot e^{x}-(x + 2)\cdot(e^{x})'}{(e^{x})^{2}}=\frac{1\cdot e^{x}-(x + 2)\cdot e^{x}}{e^{2x}}=\frac{e^{x}-(x + 2)e^{x}}{e^{2x}}$.

Step4: Simplify $f'(x)$

Factor out $e^{x}$ in the numerator: $f'(x)=\frac{e^{x}(1-(x + 2))}{e^{2x}}=\frac{-x - 1}{e^{x}}$.

Step5: Evaluate $f'(2)$

Substitute $x = 2$ into $f'(x)$: $f'(2)=\frac{-2 - 1}{e^{2}}=-\frac{3}{e^{2}}$.

Answer:

$-\frac{3}{e^{2}}$