Yet another calculus question :) Given for , find the equation of the tangent line to y where the absolute value of the slope is minimized.

I have tried taking both the first and second derivatives and setting them equal to 0 and using that as the answer, but they're incorrect. Could somebody please explain how to complete the question correctly? Thank you so much!

 y = (-4/3)*sqrt(2/3)

This is the same as writing

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Explanation:

The phrasing "where the absolute value of the slope is minimized" is an interesting way of saying "the tangent slope is 0". This is because absolute values are never negative, so the smallest it can get is 0.

Your teacher has given you

y = x^3 - 2x

which differentiates into

dy/dx = 3x^2 - 2

after using the power rule

The derivative function lets us determine the slope of the tangent. The slope is the dy/dx value. Since we want a slope of 0, we'll set 3x^2-2 equal to zero and solve for x. So you have the correct idea, but you won't involve the second derivative.

dy/dx = 0

3x^2 - 2 = 0

3x^2 = 2

x^2 = 2/3

x = sqrt(2/3)

Notice how I'm ignoring the negative version of this root. This is due to the fact that

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Now plug this x value back into the original equation to find its corresponding y coordinate.

y = x^3 - 2x

y = x(x^2 - 2)

y = sqrt(2/3)*( 2/3 - 2 )

y = sqrt(2/3)*( -4/3 )

y = (-4/3)*sqrt(2/3)

Note that x = sqrt(2/3) leads to x^2 = 2/3 after squaring both sides.

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Therefore, the equation of this tangent line is y = (-4/3)*sqrt(2/3)

All horizontal lines are of the form y = k, for some constant k. This constant value is basically what number you want the horizontal line to go through on the y axis. That number would be (-4/3)*sqrt(2/3).

my answer is$ 12

step-by-step explanation:

because 15 -2=12