Derivative absolute value and also curve help

In summary, it seems that the points on the curve y=(x+1) ³(1-3x) ² where the slope of the tangent line is horizontal are -1 , 1/3 , and -1/5 .
  • #1
MrRottenTreats
18
0
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)
 
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  • #2
MrRottenTreats said:
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 [itex] x \rightarrow x^- [/itex] ) 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

:confused: 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.
 
  • #3
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
 
  • #4
okay and for the first one you are saying that i should break it up into 2 parts

the first being:
lim [itex] x \rightarrow x^- [/itex]

so i would get x/-x
=-1

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

? is this what you are referring too?
 
  • #5
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

[tex]\lim_{x \to a}\frac{f(x)-f(a)}{x-a}[/tex] exists.

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

[tex]\lim_{x \to a^-} g(x) = \lim_{x \to a^+}g(x)[/tex].

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?
 
Last edited:
  • #6
MrRottenTreats said:
okay and for the first one you are saying that i should break it up into 2 parts

the first being:
lim [itex] x \rightarrow x^- [/itex]

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.
 
  • #7
What the hell is [tex]\lim x \to x^-[/tex]?

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.

:)
 
Last edited:

1. What is a derivative?

A derivative is a mathematical concept that represents the rate of change of a function. In other words, it shows how much a function is changing at a specific point.

2. How do you find the derivative of an absolute value function?

To find the derivative of an absolute value function, you can use the definition of a derivative and apply it to both the positive and negative portions of the function. This will result in two separate equations that can be simplified to find the derivative at any point.

3. What is the relationship between a derivative and a curve?

The derivative of a curve represents the slope of the tangent line at any given point on the curve. In other words, it shows how steep or flat the curve is at that point. The derivative also helps to determine the concavity of the curve, whether it is increasing or decreasing, and the maximum or minimum points on the curve.

4. How can derivatives be applied in real-world situations?

Derivatives have many real-world applications, such as in economics, physics, and engineering. For example, in economics, derivatives can be used to analyze the rate of change in supply and demand, while in physics, derivatives can be used to calculate the velocity and acceleration of an object. In engineering, derivatives are used to design and optimize systems and structures.

5. What are some common techniques for solving derivative problems?

Some common techniques for solving derivative problems include the power rule, product rule, quotient rule, and chain rule. These rules provide a systematic way of finding the derivative of different types of functions. Additionally, graphing and using calculus software can also help in visualizing and solving derivative problems.

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