question 12 (mandatory) (1 point)\nfind $f(1)$ if $f(x)=(sqrt{x}+1)(x^{2}-1)$.\na) 4\nb) -4\nc) 1\nd)…

Explanation:

Step1: Use product - rule

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Let $u=\sqrt{x}+1=x^{\frac{1}{2}}+1$ and $v = x^{2}-1$. Then $u'=\frac{1}{2}x^{-\frac{1}{2}}$ and $v' = 2x$.

Step2: Calculate $f'(x)$

$f'(x)=u'v + uv'=\frac{1}{2}x^{-\frac{1}{2}}(x^{2}-1)+(x^{\frac{1}{2}} + 1)\cdot2x$.

Step3: Simplify $f'(x)$

[ \begin{align*} f'(x)&=\frac{1}{2}x^{-\frac{1}{2}}\cdot x^{2}-\frac{1}{2}x^{-\frac{1}{2}}+2x^{\frac{3}{2}}+2x\ &=\frac{1}{2}x^{\frac{3}{2}}-\frac{1}{2}x^{-\frac{1}{2}}+2x^{\frac{3}{2}}+2x\ &=\left(\frac{1}{2}+2\right)x^{\frac{3}{2}}-\frac{1}{2}x^{-\frac{1}{2}}+2x\ &=\frac{5}{2}x^{\frac{3}{2}}-\frac{1}{2}x^{-\frac{1}{2}}+2x \end{align*} ]

Step4: Evaluate $f'(1)$

Substitute $x = 1$ into $f'(x)$: [ \begin{align*} f'(1)&=\frac{5}{2}(1)^{\frac{3}{2}}-\frac{1}{2}(1)^{-\frac{1}{2}}+2(1)\ &=\frac{5}{2}-\frac{1}{2}+2\ &=2 + 2\ &=4 \end{align*} ]

Answer:

A. 4