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.
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.
To answer these questions, we need to analyze the given graph and use the properties of integrals and derivatives.
1. Let's start by understanding the function g(x). The graph represents the area under the curve of f(x) from -4 to x. To find g(-1), g'(-1), and g''(-1), we need to evaluate the corresponding values.
- To find g(-1), we need to find the area under the curve of f(x) from -4 to -1. We can do this by calculating the definite integral of f(x) from -4 to -1. The value of g(-1) represents the area under the curve from -4 to -1.
- To find g'(-1), we need to find the derivative of the function g(x) and then evaluate it at x = -1.
- To find g''(-1), we need to find the second derivative of the function g(x) and then evaluate it at x = -1.
2. To find the x-coordinate of each point of inflection of the graph of g on the open interval -4 < x < 3, we need to examine the concavity of the function g(x). A point of inflection occurs when the concavity changes. This can be identified using the second derivative of g(x).
3. Let's consider the function h(x). It represents the area under the curve of f(x) from x to 3. We want to find all values of x in the closed interval -4 ≤ x ≤ 3 for which h(x) = 0. This means we need to find the points where the area under the curve from x to 3 is equal to zero. We can determine these points by calculating the definite integral of f(x) from x to 3 and solving for x when the integral equals zero.
4. Finally, we need to find all intervals on which h is decreasing. To do this, we need to analyze the behavior of the function h(x) and examine when its derivative is negative. If the derivative of h(x) is negative on an interval, then h(x) is decreasing on that interval.
To obtain the exact numerical values and detailed explanations for these questions, please provide the actual graph in a suitable format or describe the characteristics of the graph, including any important points or shape of the curve.