Looking at the graph,
g'(x) = 1/3 x - 5/3 for -4 < x < -1
g'(x) = 2x for -1 < x < 1
g(x) = ∫[-4,x] 1/3 x - 5/3 dx
for -4 <= x <= -1
g(x) = 1/6 x^2 - 5/3 x (-4,x)
= (1/6 x^2 - 5/3 x) - (16/6 + 20/3)
= 1/6 x^2 - 5/3 x - 28/3
g(x) = g(-1) + ∫[-1,1] 2x dx for -1<x<1
g(x) = g(1) + ∫[1,3] 4-2x dx for 1<x<3
The graph shown is g'(x), and is continuous, so g'(1) exists.
g(-1) exists because g' is continuous
g''(-1) does not exist
This is kind of nasty in the details, but as long as you break things up into intervals, it works out ok. Kind of a sneaky way to turn one problem into three.
Here is the graph. h t t p : / /goo.gl/PTc2I (spaces added at the beginning so it could be added as a website)
1. Let g be the function given by g(x)=integrate from -4 to x f(t)dt. For each of g(-1), g'(-1), and g''(-1), find the value of state that it does not exist.
2. For the function g defined in part 1., find the x-coordinate of each point of inflection of the graph of g on the open interval -4<x<3. Explain your reasoning.
3. Let h be the function given by h(x)= integrate from x to 3 f(t)dt. Find all values of x in the closed interval -4≤x≤3 for which h(x)=0.
4. For the function h defined in part 3, find all intervals on which h is decreasing. Explain your reasoning.
1 answer
1 answer