calc problem

[lulln]

find the critical numbers for f(x)=e^x(x^2-x)
Hi, I'm having trouble on how to start.

 

[lev888]

find the critical numbers for f(x)=e^x(x^2-x)
Hi, I'm having trouble on how to start.

What are critical numbers for a function?

 

[MarkFL]

By critical numbers, do you mean those values of \(x\) for which the derivative of the given function is zero?

 

[lulln]

Start with definition of critical numbers.

According to that definition/s, how many critical numbers you would have here?

I would say 2?

 

[lulln]

By critical numbers, do you mean those values of \(x\) for which the derivative of the given function is zero?

yes

 

[Deleted member 4993]

I would say 2?

How did you calculate that? Please share your work.

 

[lulln]

How did you calculate that?

do I take derivative and set it to 0?

 

[MarkFL]

yes

Have you computed the first derivative of the given function? What did you get?

 

[lulln]

Have you computed the first derivative of the given function? What did you get?

e^xx+e^xx^2-e^x

 

[MarkFL]

We are given:

[MATH]f(x)=e^x(x^2-x)[/MATH]
Hence:

[MATH]f'(x)=e^x(x^2-x)+e^x(2x-1)=e^x(x^2+x-1)[/MATH]
This agrees with your result. So we equate this to zero:

[MATH]f'(x)=e^x(x^2+x-1)=0[/MATH]
What do you get when equating each factor in turn to zero?

 

[lulln]

We are given:

[MATH]f(x)=e^x(x^2-x)[/MATH]
Hence:

[MATH]f'(x)=e^x(x^2-x)+e^x(2x-1)=e^x(x^2+x-1)[/MATH]
This agrees with your result. So we equate this to zero:

[MATH]f'(x)=e^x(x^2+x-1)=0[/MATH]
What do you get when equating each factor in turn to zero?

e^x=0 is no solution

 

[MarkFL]

e^x=0 is no solution

Correct. What about the other factor?

 

[lulln]

Correct. What about the other factor?

non factorable

 

[MarkFL]

non factorable

Are there other ways to find the roots of a quadratic, besides factoring?

 

[lulln]

Are there other ways to find the roots of a quadratic, besides factoring?

add and subtract 1/4 both sides?
x^2+x-1=0
(x^2+x+1/4)-1/4-1=0

 

[MarkFL]

add and subtract 1/4 both sides?
x^2+x-1=0
(x^2+x+1/4)-1/4-1=0

Well, you could complete the square, but why not just use the quadratic formula?

 

[lulln]

Well, you could complete the square, but why not just use the quadratic formula?

all coefficients will be 1?

 

[lulln]

all coefficients will be 1?

also, it's -1+/- sqrt 5/2 quadratic formula is faster thanks

 

[MarkFL]

also, it's -1+/- sqrt 5/2 quadratic formula is faster thanks

Yes, so your critical values are:

[MATH]x=\frac{-1\pm\sqrt{5}}{2}[/MATH]

 

[MarkFL]