Determining Intervals 2

[Michael13]

Thanks for the help!!

Determine the intervals on which the following function is increasing and decreasing and classify each of the critical points as a relative minimum, maximum, or neither.

g(x)= (5x^2)/(x-7)

 

[Quaid]

Determine the intervals on which the following function is increasing and decreasing and classify each of the critical points as a relative minimum, maximum, or neither.

g(x)= (5x^2)/(x-7)

Hi Michael:

We don't do homework assignments. But, if you show us your work thus far, we will help you to finish.

What did you get, when you differentiated g(x) ?

Have you studied the First Derivative Test for determining where a function is increasing or decreasing?

Can you explain why you're stuck?

Cheers

 

[Michael13]

Sorry Everyone.

Sorry all, I just wanted to throw them on here real quick. This is what I have so far:

I found g'=(5x(x-14))/((x-7)
This gave me critical points of 0 and 14
I believe that the minimum is 14 and the maximum is 0 because 7 is not in the domain and I am looking at two curves.
I believe that the function is increasing on (infinity, 0)(14, infinity) and decreasing (0,7)(7,14)
tell what you think and thank you so much!

 

[HallsofIvy]

Sorry all, I just wanted to throw them on here real quick. This is what I have so far:

I found g'=(5x(x-14))/((x-7)

This is wrong but may be a typo- the denominator should be squared.

This gave me critical points of 0 and 14
I believe that the minimum is 14 and the maximum is 0 because 7 is not in the domain and I am looking at two curves.
I believe that the function is increasing on (infinity, 0)(14, infinity) and decreasing (0,7)(7,14)
tell what you think and thank you so much!

Yes. The denominator is always positive so the sign of g' depends on the sign of the numerator. For x< 0, the numerator is the product of one positive number (5) and two negative numbers so g' is positive. For 0< x< 14, the numerator is the product of two positive and one negative number and so g' is negative. (Again, because x- 7, in the denominator, is squared, the sign does not change at x= 7). Finally, for x> 14, the numerator is the product of three positive numbers so g' is positive.

(You mean "(-infinity, 0)" not "(infinity, 0)".)

 

[Michael13]

Awesome, thank you. It is nice to know I was on the right track and thank you for catching those typos!