tmagloire1
  • tmagloire1
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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freckles
  • freckles
have you tried seeing if you can apply mean value theorem to question 1 ?
tkhunny
  • tkhunny
Mean Value Theorem... What is the slope of the extreme line segment so defined? f(−1) = −3 and f(4) = 12
tmagloire1
  • tmagloire1
I understand that i tis mean value theorem but i didn't get the same answer as them

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tmagloire1
  • tmagloire1
ill try it again 1 sec
tmagloire1
  • tmagloire1
The slope is 3 @tkhunny
freckles
  • freckles
that is right
freckles
  • 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: \[f(x)=-(x^2-10)(x^2+1)\]
tmagloire1
  • tmagloire1
I'm totally confused. I'm sorry
freckles
  • freckles
on what to do next? or what I said?
freckles
  • 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|
freckles
  • freckles
so \[|x^2-10|=x^2-10 \text{ on } (-\infty,-\sqrt{10}] \cup [\sqrt{10},\infty) \\ \text{ or } \\ |x^2-10|=-(x^2-10) \text{ on } (-\sqrt{10},\sqrt{10})\]
freckles
  • freckles
so on [0,2.5] our function looks like \[f(x)=-(x^2-10)(x^2+1)\]
freckles
  • freckles
now apply mean value theorem
freckles
  • 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)
freckles
  • freckles
and then finally solve the equation for c
tmagloire1
  • tmagloire1
I got 17.18/2.5
freckles
  • freckles
hmm... I'm getting 17.1875 on the top maybe I did something wrong or did you approximate?
freckles
  • 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
freckles
  • freckles
you need to find f'
tmagloire1
  • tmagloire1
Wait so if f'(c) =6.875 what would use evaluate for the left side? @freckles
freckles
  • freckles
the function f but you have to find derivative of f to find f'
freckles
  • freckles
have you differentiated the function on [0,2.5] yet!
tmagloire1
  • tmagloire1
18x-4x^3
freckles
  • freckles
\[f'(c)=6.875 \\ 18c-4c^3=6.875\]
tmagloire1
  • tmagloire1
ohh ok and just solve for c
freckles
  • 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]
tmagloire1
  • tmagloire1
I got 2 roots
freckles
  • 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
tmagloire1
  • tmagloire1
one last thing, for the first problem. was the answer 3 for a differentiated function or for the normal answer.
freckles
  • 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)?
tmagloire1
  • 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
freckles
  • freckles
it's cool

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