posted by cher .
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.
i got a but how do you do b and c?
if f'(x) = (4-x)^-3
then f(x) = (1/2)(4-x)^-2 + c
given f(1) = 2
2 = (1/2)(4-1)^-2 + c
2 = (1/2)(1/9) + c
c = 2 - 1/18 = 35/18
f(x) = (1/2)(4-x)^-2 + 35/18
f''(x) = -3(4-x)^-4 (-1) = 3/(4-x)^4
the graph is concave up when f''(x) > 0
the graph is concave down when f''(x) < 0
since (4-x)^4 is always positive, for x≠4
then 3/(4-x)^4 is always positive.
So the curve is concave up for all x's in the domain.
here is what Wolfram thinks of your function