# calculus

posted by on .

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 - ,

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 - ,

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 - ,

how did you find f(x)? i didn't know how to do it??