Question 1142315: Calculus Question
Consider the function  .
Use the first derivative to determine where the function is increasing or decreasing.
Find the local maximum and minimum values. Show your work.
Determine the intervals where the function is concave up or concave down.
Provide a sketch of the graph below. Label intercepts and critical points.
Answer by greenestamps(13198)   (Show Source):
You can put this solution on YOUR website!
f(x) = 3x^5-5x^3+3
f'(x) = 15x^4-15x^2 = 15x^2(x^2-1) = 15(x^2)(x-1)(x+1)
f''(x) = 60x^3-30x = 30x(x^2-1) = 30(x)(x-1)(x+1)
The critical points for both first and second derivatives are -1, 0, and 1; the resulting intervals are (-infinity, -1), (-1,0), (0,1), and (1, infinity).
An analysis, using the factored forms of the first and second derivatives and test points in each interval:
interval f'(x) f''(x)
---------------------------------------------------------------------------------
(-infinity, -1) (+)(+)(-)(-) = + increasing (+)(-)(-)(-) = - concave down
(-1,0) (+)(+)(+)(-) = - decreasing (+)(-)(+)(-) = + concave up
(0,1 ) (+)(+)(+)(-) = - decreasing (+)(+)(+)(-) = - concave down
(1,infinity) (+)(+)(+)(+) = - increasing (+)(+)(+)(+) = + concave up
A graph confirming the preceding analysis:
NOTE: While most resources use test points in each interval to determine increasing/decreasing and concave up/concave down, there is an easier and faster way to make those determinations.
Consider "walking" along the number line and seeing what happens to the signs of the factors of the derivatives each time you pass one of the critical points.
First derivative test for increasing/decreasing:
For large negative values of x, f'(x) = 15(x^2)(x-1)(x+1) is positive; the function is increasing.
When we pass x=-1, one factor in f'(x) changes sign, so the sign of f'(x) changes; the function is now decreasing.
When we pass x=0, TWO factors change sign, resulting in no change to the sign of f'(x); the function is still decreasing.
When we pass x=1, one factor changes sign, so the sign of f'(x) changes; the function is now increasing.
Summary: increasing on (-infinity,-1) and (1,infinity); decreasing on (-1,0) and (0,1).
Second derivative test for concave up/concave down:
For large negative values of x, f''(x) = 30(x)(x-1)(x+1) is negative; the function is concave down.
When we pass each of the three critical points, one factor in f''(x) changes sign, so the concavity changes at each critical point.
Summary: concave down on (-infinity,-1) and (0,1); concave up on (-1,0) and (1,infinity).
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