Question

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Answer 1

What are these 'corners' denoting?
(sorry)

Answer 2

i bet it is absolute value

Answer 3

but i could certainly be wrong

Answer 4

yeah im not really sure but it looks like absolute value

Answer 5

\(\large\color{#000000 }{ \displaystyle f(x)=\left|(x^2 - 6)(x^2 + 2)\right| }\)

First, you will need to find the slope of the secant:
\(\large\color{#000000 }{ \displaystyle \frac{ f(2)-f(1) }{2-1} }\)

Answer 6

let's say slope of secant was "c".
Then you would find f'(x), and set f'(x)=c to solve for x.

Answer 7

I will show and prove the formula for absolute value derivatives, and you while I do this, find this slope.

Answer 8

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;

\(\large\color{#000000 }{ \displaystyle \frac{d}{dx}|y|=\frac{y}{|y|}\times y' }\)

Answer 9

So, the formula you would need to apply for the derivative of your function f(x), is:

\(\Large{\bbox[5pt, lightyellow ,border:2px solid blue ]{ \displaystyle \frac{d}{dx}|y|=\frac{y}{|y|}\times y' }}\)

Answer 10

is the derivative f'(x) = (4 x (x^2-6) (x^2-2))/(abs(x^2-6))

Answer 11

\(\large\color{#000000 }{ \displaystyle f(x)=\left|(x^2 - 6)(x^2 + 2)\right| }\)

\(\large\color{#000000 }{ \displaystyle f'(x)=\frac{(x^2 - 6)(x^2 + 2)}{\left|(x^2 - 6)(x^2 + 2)\right|} \times\left[ 2x(x^2 - 6)+2x(x^2 + 2)\right]}\)

\(\large\color{#000000 }{ \displaystyle f'(x)=\frac{(x^2 - 6)(x^2 + 2)}{\left|(x^2 - 6)(x^2 + 2)\right|} \times\left[ 2x(x^2 - 6+x^2 + 2)\right]}\)

\(\large\color{#000000 }{ \displaystyle f'(x)=\frac{(x^2 - 6)(x^2 + 2)}{\left|(x^2 - 6)(x^2 + 2)\right|} \times\left[ 2x(2x^2 - 4)\right]}\)

\(\large\color{#000000 }{ \displaystyle f'(x)=\frac{(x^2 - 6)(x^2 + 2)}{\left|(x^2 - 6)(x^2 + 2)\right|} \times\left[ 4x(x^2 - 2)\right]}\)

and then I will kaboom blam this :)

Answer 12

\(\large\color{#000000 }{ \displaystyle f'(x)=\frac{(x^2 - 6)(x^2 + 2)}{\left|(x^2 - 6)(x^2 + 2)\right|} \times\left[ 4x(x^2 - 2)\right]}\)

\(\large\color{#000000 }{ \displaystyle f'(x)=\frac{(x^2 - 6)(x^2 + 2)}{\left|(x^2 - 6)\right|\cdot \left|(x^2 + 2)\right|} \times\left[ 4x(x^2 - 2)\right]}\)

|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


\(\large\color{#000000 }{ \displaystyle f'(x)=- 4x(x^2 - 2)}\)

Answer 13

kaboom like promised.

Answer 14

have you found the slope of the secant?

Answer 15

I got 0 but thats probably wrong rigt?

Answer 16

\(\large\color{#000000 }{ \displaystyle \frac{ f(2)-f(1) }{2-1} }\)

\(\large\color{#000000 }{ \displaystyle \frac{ |(2^2-6)(2^2+2)|-|(1^2-6)(1^2+2)| }{2-1} }\)

\(\large\color{#000000 }{ \displaystyle \frac{ |(-2)(6)|-|(-5)(3)| }{2-1} }\)

\(\large\color{#000000 }{ \displaystyle \frac{ 12-15 }{2-1}=-3 }\)

Answer 17

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....

Answer 18

So we are setting
f'(x)=c

Answer 19

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] ... )

Answer 20

ok I got 2 number that satisfy the conclusion of the mean value theorem