Question 864867: Hello,
I'm taking Calculus online and have a homework problem I'm having trouble with. The homework has an example that can help us, but I simply cannot put my question to the example together.
Where I got stuck was perhaps a simple step to math gurus, finding the value of x when the equation equals zero. I've done similar problems, but it's this step that always gives me the most trouble. It's just fractions - never been too great with fractions.
I'm trying to find the relative maximum and minimum point(s):
The function is as follows:
F(x)= 3-(8+7x)^2/7
Derivative:
(from the example problem, figured it out, but don't really know how the 7 cancels out)
F'(x)=-2/7(8+7x)^-5/7
F'(x)=-2(8+7x)^-5/7 Derivative.
This is the second step I also get stuck in - Solve the equation 0=-2(8+7x)^-5/7 to determine where F'(x) does not exist.
I got this far:
0/-2=(8+7x)^-5/7
0^2=(8+7x)^-5/7
0=(8+7x)^-5/7
Usually I would multiply by an exponent that would make the -5/7=1, for example if it were 1/3 it'd multiply by exponent of 3 to = 1. on both sides.
Help?
Once I figure this step I am able to input it into the F(x) equation to find out the relative maximum and relative minimum points.
Thank you! :)
Rosy
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! F(x)= 3-(8+7x)^2/7
Derivative:
(from the example problem, figured it out, but don't really know how the 7 cancels out)
F'(x)=-2/7(8+7x)^-5/7
F'(x)=-2(8+7x)^-5/7 Derivative.
---------------
The 7's cancel because d/dx of (8+7x) is 7.
---
f'(x) = 0
-2(8+7x)^-5/7 = 0
(8+7x)^-5/7 = 0
8+7x = 0
x = -8/7
-----------------
The point (-8/7,3) is a cusp, there is no max or min of the function.
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