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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
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

This Question is Closed

freckles
 one year ago
Best ResponseYou've already chosen the best response.2have you tried seeing if you can apply mean value theorem to question 1 ?

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.0Mean Value Theorem... What is the slope of the extreme line segment so defined? f(−1) = −3 and f(4) = 12

tmagloire1
 one year ago
Best ResponseYou've already chosen the best response.0I understand that i tis mean value theorem but i didn't get the same answer as them

tmagloire1
 one year ago
Best ResponseYou've already chosen the best response.0ill try it again 1 sec

tmagloire1
 one year ago
Best ResponseYou've already chosen the best response.0The slope is 3 @tkhunny

freckles
 one year ago
Best ResponseYou've already chosen the best response.2so question 2 hint: x^2+1>0 for all x x^210<0 for x between sqrt(10) and sqrt(10) so x^210=(x^210) 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: \[f(x)=(x^210)(x^2+1)\]

tmagloire1
 one year ago
Best ResponseYou've already chosen the best response.0I'm totally confused. I'm sorry

freckles
 one year ago
Best ResponseYou've already chosen the best response.2on what to do next? or what I said?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[(x^210)(x^2+1)=x^210 \cdot x^2+1=x^210 \cdot (x^2+1) \text{ for all } x \\ \text{ since } x^2+1 \text{ is positive for all } x \\ \\ \text{ now let } h(x)=x^210\] we see that graph looks like: dw:1441063542991:dw notice the part that falls under the xaxis... we need to reflect that little part to find the graph of p(x)=x^210 dw:1441063593897:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.2so \[x^210=x^210 \text{ on } (\infty,\sqrt{10}] \cup [\sqrt{10},\infty) \\ \text{ or } \\ x^210=(x^210) \text{ on } (\sqrt{10},\sqrt{10})\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2so on [0,2.5] our function looks like \[f(x)=(x^210)(x^2+1)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2now apply mean value theorem

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[f'(c)=\frac{f(2.5)f(0)}{2.50}\] 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)

freckles
 one year ago
Best ResponseYou've already chosen the best response.2and then finally solve the equation for c

freckles
 one year ago
Best ResponseYou've already chosen the best response.2hmm... I'm getting 17.1875 on the top maybe I did something wrong or did you approximate?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[f'(c)=\frac{17.1875}{2.5} \\ f'(c)=6.875\] ok but you still need to work with the left hand side of the equation

tmagloire1
 one year ago
Best ResponseYou've already chosen the best response.0Wait so if f'(c) =6.875 what would use evaluate for the left side? @freckles

freckles
 one year ago
Best ResponseYou've already chosen the best response.2the function f but you have to find derivative of f to find f'

freckles
 one year ago
Best ResponseYou've already chosen the best response.2have you differentiated the function on [0,2.5] yet!

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[f'(c)=6.875 \\ 18c4c^3=6.875\]

tmagloire1
 one year ago
Best ResponseYou've already chosen the best response.0ohh ok and just solve for c

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[4c^3+18c6.875=0\] well you can solve for c but you only need to state the amount of solutions to this equation on [0,2.5]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2so you could graph 4x^3+18x6.875 on your calculator and see how many times it crosses the xaxis on [0,2.5] and yeah should be 2 that is what i see too

tmagloire1
 one year ago
Best ResponseYou've already chosen the best response.0one last thing, for the first problem. was the answer 3 for a differentiated function or for the normal answer.

freckles
 one year ago
Best ResponseYou've already chosen the best response.2you do know the meanvalue 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)?

tmagloire1
 one year ago
Best ResponseYou've already chosen the best response.0Oh 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
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