Q. 63 - 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.

$f\text{'}\left(x\right)=5{\left(x^2+1\right)}^4\left(2x\right),f\left(0\right)=1$

Step-by-Step Solution

Answer

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

Step 1. Given Information

The given derivative is $f\text{'}\left(x\right)=5{\left(x^2+1\right)}^4\left(2x\right),f\left(0\right)=1$

$$\begin{gathered} \frac{d}{dx}f\left(x\right)=5{\left(x^2+1\right)}^4\left(2x\right) \\ df\left(x\right)=5{\left(x^2+1\right)}^4\left(2x\right)dx \end{gathered}$$

Step 2. Finding antiderivative

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

Step 3. Finding the value of C

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

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

The function become

 $$\begin{gathered} f\left(x\right)={\left(x^2+1\right)}^5+0 \\ f\left(x\right)={\left(x^2+1\right)}^5 \end{gathered}$$