Posted by **cher** on Sunday, February 26, 2012 at 7:10pm.

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?

- calculus -
**Reiny**, Sunday, February 26, 2012 at 7:27pm
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

http://www.wolframalpha.com/input/?i=%281%2F2%29%284-x%29%5E-2+%2B+35%2F18

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