Q. 2 - Calculus

Question

Critical points: 

Find the critical points of each of the following
functions.

$$\begin{gathered} f\left(x\right)=\frac{3x-2}{x-1} \\ f\left(x\right)=\frac{1}{1+\sqrt{x}} \\ f\left(x\right)=\sin \left(\frac{\mathrm{\pi }}{2}x\right) \\ f\left(x\right)=e^x(x-2) \end{gathered}$$

Step-by-Step Solution

Answer

The critical point of the function $f\left(x\right)=\frac{3x-2}{x-1}$ is $x=1$.

The critical point of the function $f\left(x\right)=\frac{1}{1+\sqrt{x}}$ is $x=0,1$.

The critical point of the function $f\left(x\right)=\sin \left(\frac{\mathrm{\pi }}{2}x\right)$ is $x=1,3,5,...$.

The critical point of the function $f\left(x\right)=e^x(x-2)$ is $x=1$.

Step 1. Given Information.

The functions are:

$$\begin{gathered} f\left(x\right)=\frac{3x-2}{x-1} \\ f\left(x\right)=\frac{1}{1+\sqrt{x}} \\ f\left(x\right)=\sin \left(\frac{\mathrm{\pi }}{2}x\right) \\ f\left(x\right)=e^x(x-2) \end{gathered}$$

Step 2. Critical points.

The critical points of the function $y=f\left(x\right)$ are values c such that $f\text{'}\left(c\right)=0 or c$ value does not exist.

Step 3. Consider the function.

Consider the function, $f\left(x\right)=\frac{3x-2}{x-1}$

$$\begin{gathered} f\text{'}\left(x\right)=\frac{(x-1)\left(3\right)-3x-2\left(1\right)}{{(x-1)}^2} \\ =-\frac{1}{{(x-1)}^2} \end{gathered}$$

$f\text{'}\left(x\right)$ does not occur at $x=1$.

Step 4. Consider the function.

Consider the function, $f\left(x\right)=\frac{1}{1+\sqrt{x}}$

$$\begin{gathered} f\text{'}\left(x\right)=-\frac{1}{{(1+\sqrt{x})}^2}\frac{1}{2\sqrt{x}} \\ {(1+\sqrt{x})}^2=1 \\ \sqrt{x}=-1 \\ x=1 \\ And\sqrt{ x}=0 \\ x=0 \end{gathered}$$

Critical points are $x=0,1$

Step 5. Determine the critical points.

Consider the function $f\left(x\right)=\sin \left(\frac{\mathrm{\pi }}{2}x\right)$,

$$\begin{gathered} f\text{'}\left(x\right)=\frac{\mathrm{\pi }}{2}\cos \left(\frac{\mathrm{\pi }}{2}x\right)=0 \\ \cos \left(\frac{\mathrm{\pi }}{2}x\right)=0 \\ \frac{ \mathrm{\pi }}{2}x=(2n+1)\frac{\mathrm{\pi }}{2}, n=1,2, 3,... \end{gathered}$$

The derivative is zero when $x=1,3,5,...$

Step 6. Determine the critical points.

Consider the function,

$$\begin{gathered} f\left(x\right)=e^x(x-2) \\ f\text{'}\left(x\right)=e^x(x-1)=0 \\ x-1=0 \\ x=1 \end{gathered}$$

The critical point is $x=1$.