Question

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

Answer 1

The relative max & min is $y=1.14$ & $y=-1.14$ 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 .

$y=4x-2tan(x)$ , Differentiating w.r.t x we get:

⇒ $\frac{d(y)}{dx}=\frac{d(4x-2tan(x))}{dx}$

⇒ $\frac{d(y)}{dx}=\frac{d(4x)-d(2tan(x))}{dx}$

⇒ $\frac{d(y)}{dx}={4-(2sec^2x)$       { $\frac{d(tanx)}{dx}=sec^2x$   }

Now, equation this to zero :

⇒ $4-(2sec^2x) = 0$

⇒ $2sec^2x = 4$

⇒ $sec^2x = 2$

⇒ $secx = \pm \sqrt{2}$

⇒ $x = sec^{-1}(\pm \sqrt{2})$

⇒ $x =(\pm \frac{\pi }{4} )$

Now , At x = $\frac{\pi}{4}$:

⇒ $y=4x-2tan(x)$

⇒ $y=4(\frac{\pi }{4})-2tan(\frac{\pi }{4})$

⇒ $y=\pi-2(1)$

⇒ $y=1.14$

At x= - $\frac{\pi}{4}$ :

⇒ $y=4x-2tan(x)$

⇒ $y=4(\frac{-\pi }{4})-2tan(\frac{-\pi }{4})$

⇒ $y=-\pi-2(-1)$

⇒ $y=-1.14$

Therefore , The relative max & min is $y=1.14$ & $y=-1.14$ respectively .

Answer 2

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,

          $\frac{dy}{dx} =\frac{d}{dx} (4\times x -2 \times tan x)$.....(1)

Equating equation (1) to Zero,

We get,

           $0 =\frac{d}{dy} (4\times x -2 \times tan x)$

           $0 = 4 -2 \times sec^{2} x$

           $4 = 2 \times sec^{2} x$

           $2 = sec^{2} x$

           $x= \frac{\pi }{4}$....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;

                   $\frac{d^{2}y }{dx^{2} } =\frac{d^{2} }{dx^{2} } (4\times x -2 \times tan x)$...(2)

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

we get,      

                   $\frac{d^{2}y }{dx^{2} } = -4$ <0      

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

by putting value of x in equation (1);

we get,

                 $y = 4\times (\pi /4 ) -2 \times tan (\pi /4 )$

                $y = \pi - 2$  .... is maximum point.

Similarly,  at  x= -pi/ 2,

                  $\frac{d^{2}y }{dx^{2} } = 1$ >0

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

by putting value of x in equation (1);

we get,

                 $y = 4\times (-\pi /2 ) -2 \times tan (-\pi /2 )$

                 $y = -2 \times \pi$.... is minimum point