Question

Please help!! calc ab

If f(x) is differentiable for the closed interval [−1, 4] such that f(−1) = −3 and f(4) = 12, then there exists a value c, −1< c < 4 such that

f '(c) = 3
f '(c) = 0
f(c) = −15
f (c) = 3

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If f(x) = ∣(x2 − 10)(x2 + 1)∣, how many numbers in the interval 0 ≤ x ≤ 2.5 satisfy the conclusion of the mean value theorem?

None
One
Two
Three

Answer 1

have you tried seeing if you can apply mean value theorem to question 1 ?

Answer 2

Mean Value Theorem... What is the slope of the extreme line segment so defined?

f(−1) = −3 and f(4) = 12

Answer 3

I understand that i tis mean value theorem but i didn't get the same answer as them

Answer 4

ill try it again 1 sec

Answer 5

The slope is 3 @tkhunny

Answer 6

that is right

Answer 7

so question 2
hint:
x^2+1>0 for all x
x^2-10<0 for x between -sqrt(10) and sqrt(10)
so |x^2-10|=-(x^2-10) on the interval [0,2.5]
so since you are only looking at [0,2.5]
the part of the function you only need to look at is:
\[f(x)=-(x^2-10)(x^2+1)\]

Answer 8

I'm totally confused. I'm sorry

Answer 9

on what to do next? or what I said?

Answer 10

\[|(x^2-10)(x^2+1)|=|x^2-10| \cdot |x^2+1|=|x^2-10| \cdot (x^2+1) \text{ for all } x \\ \text{ since } x^2+1 \text{ is positive for all } x \\ \\ \text{ now let } h(x)=x^2-10\]
we see that graph looks like:

notice the part that falls under the x-axis...
we need to reflect that little part to find the graph of p(x)=|x^2-10|