Asked by Anonymous
How do I find local minimums based on f'(x) with line segments?
I have a picture of a graph with a the points (-4,0), (-3,2.5), (0,0), (2,2), and (5,0). Here is the question:
The graph of f ' (x), the derivative of f(x), is continuous for all x and consists of five line segments as shown below. Given f (5) = 10, find the absolute minimum value of f (x) over the interval [0, 5].
f' never crosses the x-axis so that means the function is always increasing, so is the minimum at x=0? And if it is, how do you find the f(x) value for x=0?
I have a picture of a graph with a the points (-4,0), (-3,2.5), (0,0), (2,2), and (5,0). Here is the question:
The graph of f ' (x), the derivative of f(x), is continuous for all x and consists of five line segments as shown below. Given f (5) = 10, find the absolute minimum value of f (x) over the interval [0, 5].
f' never crosses the x-axis so that means the function is always increasing, so is the minimum at x=0? And if it is, how do you find the f(x) value for x=0?
Answers
Answered by
bobpursley
f'(x)=zero at x=-4,0,5. You are given max at x=5, so min has to be at x=0
now, f(x)=integral of f'(x) which is the area under the curve. So calculate the area under the curve from x=0 to 5, and subtract it from 10. In my head, I get about f(0)=6.5
now, f(x)=integral of f'(x) which is the area under the curve. So calculate the area under the curve from x=0 to 5, and subtract it from 10. In my head, I get about f(0)=6.5
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