Let f be the function given by f(x) = x^3 - 5x^2 + 3x +k, where k is a constant

a.) On what intervals is f increasing?
b.) On what intervals is the graph of f concave downward?
c.) Find the value of k for which f has 11 as its relative minimum

1)To find what intervals f is increasing take the derivative of f(x):
f'(x) = 3x^2 - 10x + 3
2) Now set that equation equal to zero and solve to find the critical numbers:
3x^2 - 10x + 3 = 0
(3x - 1) (x - 3) = 0
x = 1/3 and 3
3) Put these x values on a number line and plug in numbers into f'(x) to determine if the slope is positive (increasing) or negative (decreasing)
f is increasing at (-infinity,1/3) and (3, positive infinity)
5) Take the second derivative of f(x):
f''(x) = 6x -10
6) Set this eqn = to zero and solve for x
x = 5/3
7) Put 5/3 on a number line and plug in numbers on either side into f''(x) to find where the graph of f is concave down
graph of f is concave down at (-infinity, 5/3)
8) There is no value of k for which f has 11 as its relative minimum because k is the y intercept and, in this case, will always be the relative maximum.

You can play around with the graph at http://my.hrw.com/math06_07/nsmedia/tool…

If it was the related rates one with the cone of sand falling off the conveyor belt, i couldnt for the life of me figure it out :( sorry

Report Abuse

a.) On what intervals is f increasing?
f ' ( x ) = 3x^2 -10x + 3 =(x - 3)(3x - 1)
f ( x) is increasing for f ' (x ) > 0 then x in (- infinity, 1/3) U ( 3, + infinity)

b.) On what intervals is the graph of f concave downward?
f ' ' ( x) = 6x - 10
f (x) concave downward for f ' ' (x) < 0 then 6x -10 < 0 or x < 5/3

c.) Find the value of k for which f has 11 as its relative minimum
f (x) has a minimum for x = 3 see point a.) of the problem
then f (3) = 27 - 45 +9 + k = 11 then k = 20