Mathematics, 25.03.2020 14:38 20guadalupee73248

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

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Answer from: rleiphart1

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

Answer from: aeverettpdzrvo

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 from: Quest

since both sides equal, there are infinitely many olutions.

Answer from: Quest

shift 4 units up, and 1 unit left.

it will translate f(x) = 3^x to g(x) = 3^(x+1) + 4

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