Hey i have 2 questions here that i have finished and i have no answers to my solutions so i would like someone to check it over and see where i went wrong , im not 100% sure on this stuff and need some help , thanks alot !(adsbygoogle = window.adsbygoogle || []).push({});

Question 1: Using the definition of a derivative, show that ƒ (x) = |x| is not differentiable at x = 0.

My soluition: ( aside; √ = square root)

I know | u | = √u²

Let x = u

ƒ(x) = y = √u²

ƒ¹(x) = ½(2u) / √u² (du/dx)

= u / |u|

= x / |x|

Since subbing in x would make this undefined , x ≠ 0

Question #2: Find the points on the curve y=(x+1) ³(1-3x) ²

For horizontal dy/dx = 0

Dy/dx = 3(x+1) ²(1-3x) ² + 2(1-3x)(-3)(x+1) ³

Dy/dx = 3(x+1) ²(1-3x) ² + (-6)(1-3x)(x+1) ³

Dy/dx = 0

0 = 3(x+1) ²(1-3x) [ (1-3x) – 2(x+1) ]

Therefore

0 = (x+1) ²

X = -1

0 = (1-3x)

X = 1/3

0 = 1-3x-2x-2

0 = -5x -1

X = -1/5

Therefor the x-values of the points on the curve are -1 , 1/3 , and -1/5 (sub them into y=(x+1) ³(1-3x) ² to get corresponding y values)

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# Homework Help: Derivative absolute value and also curve help !

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