Q. 66 - Calculus

Question

In Exercises 63–68, find a function that has the given derivative and value. In each case you can find the answer with an educated guess-and-check process while thinking about the

chain rule.

$f\text{'}\left(x\right)=-6x{\left(x^2+1\right)}^4,f\left(0\right)=1$

Step-by-Step Solution

Answer

The function is $f\left(x\right)=-\frac{3}{5}{\left(x^2+1\right)}^5+\frac{8}{5}$

Step 1. Given Information

The given derivative is $f\text{'}\left(x\right)=-6x{\left(x^2+1\right)}^4,f\left(0\right)=1$

Step 2. Finding antiderivative

$$\begin{gathered} f\left(x\right)=\int -6x{\left(x^2+1\right)}^{4}dx \\ f\left(x\right)=\int -6x{\left(t\right)}^4\frac{dt}{2x} \\ f\left(x\right)=-3\int t^{4}dt \\ f\left(x\right)=-3\frac{t^5}{5}+C \\ f\left(x\right)=-\frac{3}{5}{\left(x^2+1\right)}^5+C \end{gathered}$$

Step 3. Finding the value of C

Given that $f\left(0\right)=1$

$$\begin{gathered} -\frac{3}{5}{\left(0+1\right)}^5+C=1 \\ -\frac{3}{5}+C=1 \\ C=1+\frac{3}{5}=\frac{8}{5} \end{gathered}$$

The function become $f\left(x\right)=-\frac{3}{5}{\left(x^2+1\right)}^5+\frac{8}{5}$