Posted by **bill** on Saturday, February 25, 2012 at 3:32pm.

Consider a differentiable functionf having domain all positive real numbers, and for which it is known that f'(x)=(4-x)^-3 for x.0

a. find the x-coordinate of the critical point f. DEtermine whether the point is a relative maximum or minomum, or neither for the fxn f. justify your answer.

b) find all intervals on which the graph of f is concave down. jusity answer.

c. given that f(1)=2 determine the fxn f.

here's my work so far

(4-x)^-3=0

4x^-3-((x^2)^-3)

4x^-3-x^-6=0

4-x=0.......x^-3

+4..+4......x=0

-x=4

x=-4

- calculus -
**Steve**, Saturday, February 25, 2012 at 3:58pm
if f' = (4-x)^-3

f'' = (-3)(4-x)^-4 (-1) = 3(4-x)^-4

f' and f'' are never zero.

There is a vertical asymptote ate x = 4.

How did you get from line 1 to line 2 of your solution?

f(x) = 1/2 (4-x)^-2 + C

2 = 1/18 + C

- calculus -
**bill**, Saturday, February 25, 2012 at 5:34pm
umm..i don't really know i i didn't know how to do it.. i don't even think i need that part anyway.....and i checked my work and it was x=4 not -4

- calculus -
**bill**, Saturday, February 25, 2012 at 5:35pm
how did you find f(x)? i didn't know how to do it??

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