## tmagloire1 one year ago 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 --------- 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

1. freckles

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

2. tkhunny

Mean Value Theorem... What is the slope of the extreme line segment so defined? f(−1) = −3 and f(4) = 12

3. tmagloire1

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

4. tmagloire1

ill try it again 1 sec

5. tmagloire1

The slope is 3 @tkhunny

6. freckles

that is right

7. freckles

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: $f(x)=-(x^2-10)(x^2+1)$

8. tmagloire1

I'm totally confused. I'm sorry

9. freckles

on what to do next? or what I said?

10. freckles

$|(x^2-10)(x^2+1)|=|x^2-10| \cdot |x^2+1|=|x^2-10| \cdot (x^2+1) \text{ for all } x \\ \text{ since } x^2+1 \text{ is positive for all } x \\ \\ \text{ now let } h(x)=x^2-10$ we see that graph looks like: |dw:1441063542991:dw| 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| |dw:1441063593897:dw|

11. freckles

so $|x^2-10|=x^2-10 \text{ on } (-\infty,-\sqrt{10}] \cup [\sqrt{10},\infty) \\ \text{ or } \\ |x^2-10|=-(x^2-10) \text{ on } (-\sqrt{10},\sqrt{10})$

12. freckles

so on [0,2.5] our function looks like $f(x)=-(x^2-10)(x^2+1)$

13. freckles

now apply mean value theorem

14. freckles

$f'(c)=\frac{f(2.5)-f(0)}{2.5-0}$ solve for c you will need first find f'(x) and then find f'(c) by replacing x with c you will also of course need to evaluate f(2.5) and f(0)

15. freckles

and then finally solve the equation for c

16. tmagloire1

I got 17.18/2.5

17. freckles

hmm... I'm getting 17.1875 on the top maybe I did something wrong or did you approximate?

18. freckles

$f'(c)=\frac{17.1875}{2.5} \\ f'(c)=6.875$ ok but you still need to work with the left hand side of the equation

19. freckles

you need to find f'

20. tmagloire1

Wait so if f'(c) =6.875 what would use evaluate for the left side? @freckles

21. freckles

the function f but you have to find derivative of f to find f'

22. freckles

have you differentiated the function on [0,2.5] yet!

23. tmagloire1

18x-4x^3

24. freckles

$f'(c)=6.875 \\ 18c-4c^3=6.875$

25. tmagloire1

ohh ok and just solve for c

26. freckles

$-4c^3+18c-6.875=0$ well you can solve for c but you only need to state the amount of solutions to this equation on [0,2.5]

27. tmagloire1

I got 2 roots

28. freckles

so you could graph -4x^3+18x-6.875 on your calculator and see how many times it crosses the x-axis on [0,2.5] and yeah should be 2 that is what i see too

29. tmagloire1

one last thing, for the first problem. was the answer 3 for a differentiated function or for the normal answer.

30. freckles

you do know the mean-value theorem if it satisfied the conditions then there c is between [-1,4] such that $f'(c)=\frac{f(4)-f(-1)}{4-(-1)}$ do you know notice it says f'(c) and not f(c)?

31. tmagloire1

Oh okay I missed that. Thank you for helping me with these problems! Sorry my internet was really slow and i lost connecivity half way through

32. freckles

it's cool