- #1
MrRottenTreats
- 18
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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 , I am not 100% sure on this stuff and need some help , thanks a lot !
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)
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)