chris215
  • chris215
.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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SolomonZelman
  • SolomonZelman
What are these 'corners' denoting? (sorry)
anonymous
  • anonymous
i bet it is absolute value
anonymous
  • anonymous
but i could certainly be wrong

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chris215
  • chris215
yeah im not really sure but it looks like absolute value
SolomonZelman
  • SolomonZelman
\(\large\color{#000000 }{ \displaystyle f(x)=\left|(x^2 - 6)(x^2 + 2)\right| }\) First, you will need to find the slope of the secant: \(\large\color{#000000 }{ \displaystyle \frac{ f(2)-f(1) }{2-1} }\)
SolomonZelman
  • SolomonZelman
let's say slope of secant was "c". Then you would find f'(x), and set f'(x)=c to solve for x.
SolomonZelman
  • SolomonZelman
I will show and prove the formula for absolute value derivatives, and you while I do this, find this slope.
SolomonZelman
  • SolomonZelman
The definition of the absolute value is: \(\large\color{#000000 }{ \displaystyle |x|=\sqrt{x^2} }\) (same for a function y) \(\large\color{#000000 }{ \displaystyle |y|=\sqrt{y^2} }\) ----------------------------- \(\large\color{#000000 }{ \displaystyle \frac{d}{dx}|y|=\frac{d}{dx}\sqrt{y^2} }\) \(\large\color{#000000 }{ \displaystyle =\frac{1}{2\sqrt{y^2}}\times 2y\times y' }\) (applying the chain rule twice) \(\large\color{#000000 }{ \displaystyle \frac{d}{dx}|y|=\frac{y}{\sqrt{y^2}}\times y' }\) recall, the defnition of the absolute value; \(\large\color{#000000 }{ \displaystyle \frac{d}{dx}|y|=\frac{y}{|y|}\times y' }\)
SolomonZelman
  • SolomonZelman
So, the formula you would need to apply for the derivative of your function f(x), is: \(\Large{\bbox[5pt, lightyellow ,border:2px solid blue ]{ \displaystyle \frac{d}{dx}|y|=\frac{y}{|y|}\times y' }}\)
chris215
  • chris215
is the derivative f'(x) = (4 x (x^2-6) (x^2-2))/(abs(x^2-6))
SolomonZelman
  • SolomonZelman
\(\large\color{#000000 }{ \displaystyle f(x)=\left|(x^2 - 6)(x^2 + 2)\right| }\) \(\large\color{#000000 }{ \displaystyle f'(x)=\frac{(x^2 - 6)(x^2 + 2)}{\left|(x^2 - 6)(x^2 + 2)\right|} \times\left[ 2x(x^2 - 6)+2x(x^2 + 2)\right]}\) \(\large\color{#000000 }{ \displaystyle f'(x)=\frac{(x^2 - 6)(x^2 + 2)}{\left|(x^2 - 6)(x^2 + 2)\right|} \times\left[ 2x(x^2 - 6+x^2 + 2)\right]}\) \(\large\color{#000000 }{ \displaystyle f'(x)=\frac{(x^2 - 6)(x^2 + 2)}{\left|(x^2 - 6)(x^2 + 2)\right|} \times\left[ 2x(2x^2 - 4)\right]}\) \(\large\color{#000000 }{ \displaystyle f'(x)=\frac{(x^2 - 6)(x^2 + 2)}{\left|(x^2 - 6)(x^2 + 2)\right|} \times\left[ 4x(x^2 - 2)\right]}\) and then I will kaboom blam this :)
SolomonZelman
  • SolomonZelman
\(\large\color{#000000 }{ \displaystyle f'(x)=\frac{(x^2 - 6)(x^2 + 2)}{\left|(x^2 - 6)(x^2 + 2)\right|} \times\left[ 4x(x^2 - 2)\right]}\) \(\large\color{#000000 }{ \displaystyle f'(x)=\frac{(x^2 - 6)(x^2 + 2)}{\left|(x^2 - 6)\right|\cdot \left|(x^2 + 2)\right|} \times\left[ 4x(x^2 - 2)\right]}\) |x^2+2| is same as just x^2+2 (for real number - we are not considering imaginaries) \(\large\color{#000000 }{ \displaystyle f'(x)=\frac{x^2 - 6}{\left|x^2 - 6\right|} \times\left[ 4x(x^2 - 2)\right]}\) x^2-6 and |x^2-6| will have the same magnitude, however, on the interval [1,2] the |x^2-6| is always negative. So, for x\(\in\)[1,2] (x^2-6) / |x^2-6| = -1 \(\large\color{#000000 }{ \displaystyle f'(x)=- 4x(x^2 - 2)}\)
SolomonZelman
  • SolomonZelman
kaboom like promised.
SolomonZelman
  • SolomonZelman
have you found the slope of the secant?
chris215
  • chris215
I got 0 but thats probably wrong rigt?
SolomonZelman
  • SolomonZelman
\(\large\color{#000000 }{ \displaystyle \frac{ f(2)-f(1) }{2-1} }\) \(\large\color{#000000 }{ \displaystyle \frac{ |(2^2-6)(2^2+2)|-|(1^2-6)(1^2+2)| }{2-1} }\) \(\large\color{#000000 }{ \displaystyle \frac{ |(-2)(6)|-|(-5)(3)| }{2-1} }\) \(\large\color{#000000 }{ \displaystyle \frac{ 12-15 }{2-1}=-3 }\)
SolomonZelman
  • SolomonZelman
By the way, the Rolle's Theorem is the same as mean value theorem, except that rolle's theorem applies specifically when the slope of the secant is 0 (and The mean Value Theorem is for all other slopes).... I don't know why the same thing for a slope of 0 has a designated name....
SolomonZelman
  • SolomonZelman
So we are setting f'(x)=c
SolomonZelman
  • SolomonZelman
can you set this up and solve for x? (the solution will give the the x-values at which the slope of the curve is the same as the slope of the secant over the interval x=[1,2] ... )
chris215
  • chris215
ok I got 2 number that satisfy the conclusion of the mean value theorem

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