Respuesta :
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,
[tex]\frac{dy}{dx} =\frac{d}{dx} (4\times x -2 \times tan x)[/tex].....(1)
Equating equation (1) to Zero,
We get,
[tex]0 =\frac{d}{dy} (4\times x -2 \times tan x)[/tex]
[tex]0 = 4 -2 \times sec^{2} x[/tex]
[tex]4 = 2 \times sec^{2} x[/tex]
[tex]2 = sec^{2} x[/tex]
[tex]x= \frac{\pi }{4}[/tex]....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;
[tex]\frac{d^{2}y }{dx^{2} } =\frac{d^{2} }{dx^{2} } (4\times x -2 \times tan x)[/tex]...(2)
putting x= pi/4 in equation (2),
we get,
[tex]\frac{d^{2}y }{dx^{2} } = -4[/tex] <0
So, we have local max at x=pi/4
by putting value of x in equation (1);
we get,
[tex]y = 4\times (\pi /4 ) -2 \times tan (\pi /4 )[/tex]
[tex]y = \pi - 2[/tex] .... is maximum point.
Similarly, at x= -pi/ 2,
[tex]\frac{d^{2}y }{dx^{2} } = 1[/tex] >0
So, we have local min at x=-pi/2
by putting value of x in equation (1);
we get,
[tex]y = 4\times (-\pi /2 ) -2 \times tan (-\pi /2 )[/tex]
[tex]y = -2 \times \pi[/tex].... is minimum point
The relative max & min is [tex]y=1.14[/tex] & [tex]y=-1.14[/tex] 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 .
[tex]y=4x-2tan(x)[/tex] , Differentiating w.r.t x we get:
⇒ [tex]\frac{d(y)}{dx}=\frac{d(4x-2tan(x))}{dx}[/tex]
⇒ [tex]\frac{d(y)}{dx}=\frac{d(4x)-d(2tan(x))}{dx}[/tex]
⇒ [tex]\frac{d(y)}{dx}={4-(2sec^2x)[/tex] { [tex]\frac{d(tanx)}{dx}=sec^2x[/tex] }
Now, equation this to zero :
⇒ [tex]4-(2sec^2x) = 0[/tex]
⇒ [tex]2sec^2x = 4[/tex]
⇒ [tex]sec^2x = 2[/tex]
⇒ [tex]secx = \pm \sqrt{2}[/tex]
⇒ [tex]x = sec^{-1}(\pm \sqrt{2})[/tex]
⇒ [tex]x =(\pm \frac{\pi }{4} )[/tex]
Now , At x = [tex]\frac{\pi}{4}[/tex]:
⇒ [tex]y=4x-2tan(x)[/tex]
⇒ [tex]y=4(\frac{\pi }{4})-2tan(\frac{\pi }{4})[/tex]
⇒ [tex]y=\pi-2(1)[/tex]
⇒ [tex]y=1.14[/tex]
At x= - [tex]\frac{\pi}{4}[/tex] :
⇒ [tex]y=4x-2tan(x)[/tex]
⇒ [tex]y=4(\frac{-\pi }{4})-2tan(\frac{-\pi }{4})[/tex]
⇒ [tex]y=-\pi-2(-1)[/tex]
⇒ [tex]y=-1.14[/tex]
Therefore , The relative max & min is [tex]y=1.14[/tex] & [tex]y=-1.14[/tex] respectively .
