Question

webwork 2 - topics 4 - 5: problem 5 (1 point) given that (f(x)=x^{10}h(x)), (h(-1)=5), (h(-1)=8). calculate (f(-1)=)

Explanation:

Step1: Apply product - rule

The product - rule states that if $f(x)=u(x)v(x)$, then $f^{\prime}(x)=u^{\prime}(x)v(x)+u(x)v^{\prime}(x)$. Here, $u(x)=x^{10}$ and $v(x)=h(x)$. So, $f^{\prime}(x)=(x^{10})^{\prime}h(x)+x^{10}h^{\prime}(x)$.
Since $(x^{n})^{\prime}=nx^{n - 1}$, then $(x^{10})^{\prime}=10x^{9}$. So, $f^{\prime}(x)=10x^{9}h(x)+x^{10}h^{\prime}(x)$.

Step2: Substitute $x = - 1$

Substitute $x=-1$ into $f^{\prime}(x)$.
We know that $h(-1) = 5$ and $h^{\prime}(-1)=8$.
When $x=-1$, $f^{\prime}(-1)=10(-1)^{9}h(-1)+(-1)^{10}h^{\prime}(-1)$.
Since $(-1)^{9}=-1$ and $(-1)^{10}=1$, we have $f^{\prime}(-1)=10\times(-1)\times5 + 1\times8$.

Step3: Calculate the value

$f^{\prime}(-1)=-50 + 8=-42$.

Answer:

$-42$