Which functions has the fastest rate of increase?

Mathematics

1 answer:

Mashcka [7]3 years ago

6 0

The function with the higgest rate of increase is 2^x

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Name the property used to make the conclusion. If x=3, then 5x=15.

bulgar [2K]

Note that both sides were multiplied by 5.

So that would be "Multiplication Property of Equality."

Basically, this property states that if a=b, then ac = bc.

6 0

4 years ago

Verify the identity cot(x-pi/2)=-tan x

vampirchik [111]

To verify the identity we need the following identities:

i) $\displaystyle{ \cot(x)= \frac{\cos x}{\sin x}$

ii) $\displaystyle{ \sin (x-y)=\ sinx\cdot\ cosy -\ siny\cdot\ cosx$

iii) $\displaystyle{ \cos (x-y)=\ cosx\cdot \ cosy +\ sinx\cdot\ siny$ .

Also, we have know that $\displaystyle{ \sin \frac{ \pi }{2}=1$ and $\displaystyle{ \cos \frac{ \pi }{2}=0.$

Thus, $\displaystyle{ \cot(x-\frac{\pi}{2})= \frac{\cos (x-\frac{\pi}{2})}{\sin (x-\frac{\pi}{2})}$

By (ii) and (iii) we have:

$\displaystyle{ \frac{\cos (x-\frac{\pi}{2})}{\sin (x-\frac{\pi}{2})}= \frac{\ cosx\cdot \ cos\frac{\pi}{2} +\ sinx\cdot\ sin\frac{\pi}{2}}{\ sinx\cdot\ cos\frac{\pi}{2} -\ sin\frac{\pi}{2}\cdot\ cosx} = \frac{\ sinx}{-\cos x}$

by simplifying $\displaystyle{ \sin \frac{ \pi }{2}=1$ and $\displaystyle{ \cos \frac{ \pi }{2}=0.$

Now, $\displaystyle{ \frac{\ sinx}{-\cos x}$ is clearly -tanx.