anonymous
  • anonymous
if f(x)=abs((x^2-12)(x^2+4)), how many numbers in the interval -2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
First, you need to put this function into something you can handle. You can use the definition of absolute value to do this:\[f(x)=|(x^2-12)(x^2+4)|=\sqrt{[(x^2-12)(x^2+4)]^2}\]so\[f(x)=\sqrt{(x^2-12)^2(x^2+4)^2}\]\[=\sqrt{x^8-16 x^6-32 x^4+768 x^2+2304}\]\[=\sqrt{(x^4-8 x^2-48)^2}\]Now, the mean value theorem says that if a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a point c in (a, b) such that\[f'(c)=\frac{f(b)-f(a)}{b-a}\]You need to find all those c's in (2,3) that satisfy,\[f'(c)=\frac{f(3)-f(2)}{3-2}=f(3)-f(2)\]...end part 1
nowhereman
  • nowhereman
only that its (-2, 3)
anonymous
  • anonymous
lol I love how you always clean up the question.

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anonymous
  • anonymous
nowhereman, do you want to finish this? I want to go eat.
nowhereman
  • nowhereman
Ok, so you are looking for the points \[c \in (-2, 3)\] so that \[f'(c) = \frac{f(3) - f(-2)}{3-(-2)} = \frac{39-64}{5} = 5\] So we need to find the solutions of \[\frac{2(x^4 - 8x^2-48)(4x^3 - 16x)}{2\sqrt{(x^4-8x^2-48)^2}} = 5\]
nowhereman
  • nowhereman
Oh and we can do it a little bit easier, because \[x \in [-2, 3] ⇒ x^2-12 < 0 ∧ x^2+4 > 0 \]\[⇒ f(x) = |(x^2-12)(x^2+4)| = -(x^2-12)(x^2+4)\]
nowhereman
  • nowhereman
So \[f'(x) = -4x^3 + 16x\] and we need to solve \[-4x^3 + 16x - 5 = 0\] by differentiating you get that the extrema of \[-4x^3 + 16x - 5\] are at \[\pm 2\sqrt{\frac{1}{3}}\]
nowhereman
  • nowhereman
the function values are there \[\pm32\frac{2}{3}\sqrt{\frac 1 3}-5\] one is bigger then zero and the other is smaller then zero. And so there must be 3 different solutions.
nowhereman
  • nowhereman
you should maybe draw the graph of f' or let some plotter do it for you, to get a better imagination of the situation.
anonymous
  • anonymous
thank you both so much for providing such a detailed answer! u have saved my calc grade!
anonymous
  • anonymous
You're welcome

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