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10 1 point Let $f(x)$ be a differentiable function for $-3 \le x \le 1$. Over this interval $f(x)$ has exactly one critical point at $x = 0$ where $f(0) = 4$. We also know that $f(-3) = -1$ and $f(1) = 2$. Which of the following must be true? The global maximum value of $f(x)$ over the interval is 4. There is no global maximum over this interval. There is a global maximum, but we cannot determine what it is based on the information given. The global minimum value of $f(x)$ over the interval is -3. Clear my selection 

          10
1 point
Let $f(x)$ be a differentiable function for $-3 \le x \le 1$. Over this interval $f(x)$ has exactly one critical point at $x = 0$ where $f(0) = 4$. We also know that $f(-3) = -1$ and
$f(1) = 2$. Which of the following must be true?
The global maximum value of $f(x)$ over the interval is 4.
There is no global maximum over this interval.
There is a global maximum, but we cannot determine what it is based on the information given.
The global minimum value of $f(x)$ over the interval is -3.
Clear my selection
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10 1 point Let f(x) be a differentiable function for -3 ≤ x ≤ 1. Over this interval f(x) has exactly one critical point at x = 0 where f(0) = 4. We also know that f(-3) = -1 and f(1) = 2. Which of the following must be true? The global maximum value of f(x) over the interval is 4. There is no global maximum over this interval. There is a global maximum, but we cannot determine what it is based on the information given. The global minimum value of f(x) over the interval is -3. Clear my selection

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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101 point Let f(x) be a differentiable function for -3 <= x <= 1. Over this interval, f(x) has exactly one critical point at x = 0 where f(0) = 4. We also know that f(-3) = -1 and f(1) = 2. Which of the following must be true? The global maximum value of f(x) over the interval is 4. There is no global maximum over this interval. There is a global maximum, but we cannot determine what it is based on the information given. The global minimum value of f(x) over the interval is -3. Clear my selection 101 point Let f(x) be a differentiable function for -3 <= x <= 1. Over this interval, f(x) has exactly one critical point at x = 0 where f(0) = 4. We also know that f(-3) = -1 and f(1) = 2. Which of the following must be true? The global maximum value of f(x) over the interval is 4. There is no global maximum over this interval. There is a global maximum, but we cannot determine what it is based on the information given. The global minimum value of f(x) over the interval is -3. Clear my selection
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Transcript

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00:01 So here we are asked to find the absolute extreme values for a few different functions on closed intervals, starting with f of x equals x squared plus 3x minus 10 on the interval from negative 6 to 2.
00:15 So the procedure here should start with evaluating the function at the end points.
00:20 So we have, let's see here, x squared plus 3x minus 10.
00:25 We'll substitute in x is negative 6 and x is 2.
00:29 So we find that f of negative 6 is equal to 8, and we find that f of 2 is equal to 0.
00:42 So those are candidates for extreme values.
00:45 Then we'll take the derivative of the given function.
00:48 So we'll have f prime of x is equal to 2x plus 3.
00:53 We'll want to solve for when this equals 0.
00:55 So that would be when x equals negative 3 over 2.
00:59 So we'll evaluate our function at negative 3 over 2.
01:02 Then and let's see here i'm just going to change that so we get a erical result so we can see then that the absolute minimum is negative 12 over 25 and the absolute maximum is 8 so for two and three you would use the exact same procedure just they are pretty simple functions to differentiate just want to find the critical points evaluate at them and compare the values there to the values on the endpoints for number four though that one is a little bit more tricky, so i'll go into that one explicitly...
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