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
[tex]f(x)=3tan^{-1}(x)-\frac{3}{2} x+5[/tex] on ([tex]-\infty,\infty[/tex])
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
[tex]f'(x)=\frac{3}{1+x^2}-\frac{3}{2}[/tex]
[tex]\frac{d(tan^{-1}x)}{dx}=\frac{1}{1+x^2}[/tex]
[tex]f'(x)=0[/tex]
[tex]\frac{3}{1+x^2}-\frac{3}{2}=0[/tex]
[tex]\frac{3}{1+x^2}=\frac{3}{2}[/tex]
[tex]1+x^2=2[/tex]
[tex]x^2=2-1=1[/tex]
[tex]x=\pm1[/tex]
Therefore, the critical points of the given function are 1 and -1.
[tex]f(0)=3-\frac{3}{2}=\frac{3}{2}[/tex]
[tex]f'(1)=0[/tex]
[tex]f'(2)=\frac{3}{5}-\frac{3}{2}=-\frac{9}{10}[/tex]
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.
[tex]f(-2)=\frac{3}{5}-\frac{3}{2}=-\frac{9}{10}[/tex]
[tex]f(-1)=0[/tex]
[tex]f(0)=\frac{3}{2}[/tex]
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 .
