Q. 1 - Calculus

Question

 Find a function f that has the given derivative f'  and value f(c), if possible. 

$f\text{'} \left(x\right) = x^8 - x^2 + 1, f\left(0\right) =- 1$

Step-by-Step Solution

Answer

The function is this $\frac{x^9}{9}-\frac{x^3}{3}+x+c$

and value of f(-1)=$\frac{-20}{9}$

Step 1.Given

The given function is $f \text{'}\left(x\right) = x^8 - x^2 + 1, f\left(0\right) =- 1$

Step 2. Finding the function using integration

$$\begin{gathered} \mathrm{Integrating} \mathrm{both} \mathrm{sides} \mathrm{with} \mathrm{respect} \mathrm{to} x, \\ \int f \text{'}\left(x\right) = \int x^8 - x^2 + 1 dx \\ =\frac{x^9}{9}-\frac{x^3}{3}+x+c \\ \mathrm{where} \mathrm{c} \mathrm{is} \mathrm{constant} \end{gathered}$$

Step 3. Finding the value of f(c).

$$\begin{gathered} when f\left(0\right)=-1 \\ f\left(x\right)=\frac{x^9}{9}-\frac{x^3}{3}+x+c \\ f\left(0\right)=0-0+0+c \\ -1=c \\ f\left(c\right) \\ =f(-1)=\frac{{(-1)}^9}{9}-\frac{{(-1)}^3}{3}+(-1)+-1 \\ =\frac{1}{9}-\frac{1}{3}-1-1 \\ =\frac{-2}{9}-2 \\ =\frac{-20}{9} \end{gathered}$$