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)=(4x)^3 for x.0
a. find the xcoordinate 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
(4x)^3=0
4x^3((x^2)^3)
4x^3x^6=0
4x=0.......x^3
+4..+4......x=0
x=4
x=4

calculus  Steve, Saturday, February 25, 2012 at 3:58pm
if f' = (4x)^3
f'' = (3)(4x)^4 (1) = 3(4x)^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 (4x)^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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