Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

i used the quotient rule!

i got my critical point 5 and 1 but it's wrong :(

-5/(x^2-2x+1)= 0

i doubt there are any

why?

the derivative is \(\frac{-5}{(x-1)^2}\)

a fraction is only zero if the numerator is zero
this numerator is \(-5\)

-5/(x^2-2x+1)= 0

right so when i get to this: -5/(x^2-2x+1)= 0 what do i do?

right

you may remember what such a thing looks like from a pre - calc course

ok but then why isn't -5 considered a critical point?

i know 1 gives you undefined.

you mean the denominator?

i did, but what i don't understand why it has no critical point.

okay but in this question im not allowed to use the calculator.

well, because it doesn't
does \(f(x)=2x+1\) have a critical point?

of course it doesn't, because for \(f(x)=2x+1\)you have a line that is always increasing

when is a fraction equal to zero?

oh no it doesn't.

and 0 does not multiply -5..

is \(\frac{5}{x}=0\) solvable?

nope.

i think i got it...

right, and neither is \(\frac{-5}{(x-1)^2}\)

thanks @satellite73 :)

how about \(e^x\)? is that ever zero?
yw

no

1?

no, \(e^x>0\) for all \(x\)
\(e^1=e\) not zero

the function \(f(x)=e^x\) is strictly increasing.

right.

@satellite73 i have a question: how come -5/(x-1)=0 is not a critical point and -4/15 is?