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Question 999910: Please check my work
The problem gives me the function:
f(x) = ((2x-1)^2)/(2x^2)
I must find the intervals where f is increasing and decreasing.
And
I must find the intervals where the graph is concave down and concave up.
And
Identify all relative extreme values of f.
Here's what I did:
f(x) = ((2x-1)^2)/(2x^2)
f'(x) = (2x-1)/(x^3)
f"(x) = (3-4x)/(x^4)
These were given already by the problem. I am worried that wouldn't be able to figure out these derivatives they seem tricky to get them simplified into those forms.
To find where f(x) is incr/decr I look at the 1st derivative and set equal to zero and get: x = 1/2 I pretty sure this exists because it's in the NUMERATOR. However, the part which I am really confused by is what to do about the denominator. Since it can never be zero how can it ever be anything. Someone told me that if it's odd then we "consider it" if it's even we "dismiss it" if this is true, why is it?
Provided it is. x =1/2 and x = 0 for critical values.
testing pts....
<--(-1)--[0]--(+1/4)--[+1/2]--(+1)-->
f'(-1) = ((2)(-1)-1)/((-1)^3) = +
f'(1/4) = (2(1/4)-1)/((1/4)^3) = -
f'(1) = (2-1)/(1^3) = +
Therefore,
is incr int(-inf,0)U(1/2,1)
is decr int(0,1/2)
For testing concave up/down intervals
I found the inflection pt from setting the 2nd derivative to zero and testing points around it.
f"(x) = 3-4x = 0
//now in this case, there was a x^4 in the denominator, but since it was even we disregard it?
x = -3/4
<--(-1)--[-3/4]--(0)-->
f"(0) = DNE because of the zeros
f"(-1) = (3-4(-1))/((-1)^4) = +
Therfore,
It is concave up (-inf,-3/4)
And concave down DNE
Please correct any errors I have made.
Thank you
Answer by Fombitz(32388)   (Show Source):
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