Question

Find the critical points of the function and use the First Derivative Test to determine whether the critical point is a local minimum or maximum (or neither). (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) f(x) = 3 tan−1(x) − 3/2x + 5, on (−[infinity], [infinity])

Answer 1

Answer:

Critical points are 1 and -1

Maximum at x=1

Minimum at x=-1

Step-by-step explanation:

We are given that a function

$f(x)=3tan^{-1}(x)-\frac{3}{2} x+5$ on ($-\infty,\infty$)

We have to find the critical points of the function.

To find the critical point we will differentiate function w.r.t x and then substitute f'(x)=0

$f'(x)=\frac{3}{1+x^2}-\frac{3}{2}$

$\frac{d(tan^{-1}x)}{dx}=\frac{1}{1+x^2}$

$f'(x)=0$

$\frac{3}{1+x^2}-\frac{3}{2}=0$

$\frac{3}{1+x^2}=\frac{3}{2}$

$1+x^2=2$

$x^2=2-1=1$

$x=\pm1$

Therefore, the critical points of the given function are 1 and -1.

$f(0)=3-\frac{3}{2}=\frac{3}{2}$

$f'(1)=0$

$f'(2)=\frac{3}{5}-\frac{3}{2}=-\frac{9}{10}$

When we goes from 0 to 2 then the sign of derivative  change from positive to negative .Therefore, function has local maximum at x=1.

$f(-2)=\frac{3}{5}-\frac{3}{2}=-\frac{9}{10}$

$f(-1)=0$

$f(0)=\frac{3}{2}$

When we goes form -2 to 0 then the sign of derivative change from negative to positive .Hence , function has local minimum at x=-1

Hence, critical points are local maximum and local minimum .