Q. 13 - Calculus

Question

Find the critical points of the function 

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

Step-by-Step Solution

Answer

$\mathrm{Therefore} \mathrm{the} \mathrm{critical} \mathrm{points} \mathrm{are} x=\pm \sqrt{\left(15\right)}$

Step 1. Given

The given function is $f^{\text{'}}\left(x\right)=\sqrt{1+x^2}-4$.

Step 2. Calculation

$$\begin{gathered} Set f\text{'}\left(x\right)=0 to determine the critical point. \\ \sqrt{1+x²}-4=0 \\ \sqrt{(1+x^2)}=4 \\ 1+x^2=16 \\ {x}^2=16-1 \\ {x}^2=15 \\ x=\pm \sqrt{\left(15\right)} \\ \mathrm{Therefore} \mathrm{the} \mathrm{critical} \mathrm{points} \mathrm{are} x=\pm \sqrt{\left(15\right)} \end{gathered}$$