Derivative absolute value and also curve help

[MrRottenTreats]

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)

 

[nrqed]

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

what you have to show is that the limit from the left (as you approach x \rightarrow x^- ) gives a different value than the limit from the right. Just use your answer for the derivative an dit will be obvious that one limit gives +1 and the other gives -1.

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

Are there some words missing in the question?? The points of inflexion? the extrema?
Looks like you are looking for the extrema.

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)

At a quick glance it seems right to me.

 

[MrRottenTreats]

hey sry i forgot the second part

Question #2: Find the point on the curve y=(x+1) ³(1-3x) ² where the slope of the tagent line is horizontal

sorry about that

 

[MrRottenTreats]

okay and for the first one you are saying that i should break it up into 2 parts

the first being:
lim x \rightarrow x^-

so i would get x/-x
=-1

and then:
lim x -> x
gettinng x/x
= 1

? is this what you are referring too?

 

[ZioX]

Your original proof for #1 is good enough. However, the question asked you to do it by the definitions.

A function is differentiable at x=a IF AND ONLY IF

\lim_{x \to a}\frac{f(x)-f(a)}{x-a} exists.

In general, limits of a function g at x=a exist IF AND ONLY IF

\lim_{x \to a^-} g(x) = \lim_{x \to a^+}g(x).

These are your definitions. Be sure to use them in your proof. Explicitly. You seem to have the right idea, but you should elaborate on your thoughts (and notation).

For #2, what is the slope of a horizontal line? So, when could the slope of the tangent be the same as this number?

 

[nrqed]

okay and for the first one you are saying that i should break it up into 2 parts

the first being:
lim x \rightarrow x^-

so i would get x/-x
=-1

and then:
lim x -> x
gettinng x/x
= 1

? is this what you are referring too?

that's right.

 

[ZioX]

What the hell is \lim x \to x^-?

Don't encourage such bad notation. He has the right idea, but hasn't put forth a good argument.

If you merely meant that his idea was right, disregard previous comment.

:)