Y=4x-2tan(x), (-pi/2,pi/2)
which are the relative min and max?

relative max  = pi -2

relative min   = -2 pi

Step-by-step explanation:

y= 4x- 2tan(x), (-pi/2,pi/2)

To find relative max and min,

According to mathematical procedure to find relative max and relative min,

the following procedure should be carry out.

Derivating equation with respect to x,

          .....(1)

Equating equation (1) to Zero,

We get,

           ....stationary point

There are total three points

1. -pi/2      2. pi/2      3. pi/4

checking local min and local max

first considering -pi/2;

                   ...(2)

putting x= pi/4 in equation (2),

we get,      

                    <0      

So, we have local max at x=pi/4

by putting value of x in equation (1);

we get,

                 .... is maximum point.

Similarly,  at  x= -pi/ 2,

                  >0

So, we have local min at x=-pi/2

by putting value of x in equation (1);

we get,

                 .... is minimum point

The relative max & min is & respectively .

Step-by-step explanation:

Here we have , y=4x-2tan(x) in interval (-pi/2,pi/2) . We need to find  the relative min and max . Let's find out;

We will first differentiate this function with respect to x and then , we'll put it to zero to find values of x for which function is maximum & minimum .

, Differentiating w.r.t x we get:

⇒

⇒

⇒       {   }

Now, equation this to zero :

⇒

⇒

⇒

⇒

⇒

⇒

Now , At x = :

⇒

⇒

⇒

⇒

At x= - :

⇒

⇒

⇒

⇒

Therefore , The relative max & min is & respectively .

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step-by-step explanation:

ian gonna lie that is way to much