• ProgMetal
In summary, the function y= |x2-4x+3| is continuous but differentiable only at the points x= 3/2 and x= 1. It has a derivative that is undefined at x= 3 and x= 1.
ProgMetal

## Homework Statement

For the function y= |x2-4x+3|
a. Sketch the function
b. Is the function continuous? Explain.
c. Is the function differentiable everywhere? explain
d. find gradient of curve at the point where x = 3/2

## Homework Equations

chain rule, differentiating...

## The Attempt at a Solution

a. easy enough.
b. I said the function was continuous as i don't think it would be undefined at any point..?
c. I know the derivative is undefined at x=3 and x=1 (the x intercepts) but I am not sure why...
d. Not sure where to go here, all i know is that i can put it in the form f(x)=|(x-3)(x-1)|, let u=x-3, let v=x-1, and that |u|=sqrt(u2), and |v|=sqrt(v2)
But I am not sure where to go from there..

ProgMetal said:

## Homework Statement

For the function y= |x2-4x+3|
a. Sketch the function
b. Is the function continuous? Explain.
c. Is the function differentiable everywhere? explain
d. find gradient of curve at the point where x = 3/2

## Homework Equations

chain rule, differentiating...

## The Attempt at a Solution

a. easy enough.
b. I said the function was continuous as i don't think it would be undefined at any point..?
c. I know the derivative is undefined at x=3 and x=1 (the x intercepts) but I am not sure why...
A derivative formula would be helpful here.
$$\frac{d|x|}{dx} = \frac{x}{|x|}$$
Notice that this reduces to 1 if x > 0, and -1 if x < 0.

Can you come up with a chain rule form of this formula? I.e.,
$$\frac{d|u|}{dx}$$

If so, that would give you a justification for saying that the derivative was undefined at x = 3 and x = 1, and you could use it to find the slope of the tangent line at x = 3/2.
ProgMetal said:
d. Not sure where to go here, all i know is that i can put it in the form f(x)=|(x-3)(x-1)|, let u=x-3, let v=x-1, and that |u|=sqrt(u2), and |v|=sqrt(v2)
But I am not sure where to go from there..

ProgMetal said:

## Homework Statement

For the function y= |x2-4x+3|
a. Sketch the function
b. Is the function continuous? Explain.
c. Is the function differentiable everywhere? explain
d. find gradient of curve at the point where x = 3/2

## Homework Equations

chain rule, differentiating...

## The Attempt at a Solution

a. easy enough.
Good!

b. I said the function was continuous as i don't think it would be undefined at any point..?
Yes, the function is continuous but that's not a very good explanation! A function does not have to be "undefined" any where in order not to be continuous. The definition of "continuous" is that $\lim_{x\to a} f(x)= f(a)$. The function f(x)= 0 if x< 0 , 1 if x>= 0 is defined for all x but not continuous at x= 0. Since you have already graphed the function what does the graph tell you about continuity? Continuity should be trivial except possibly at two points. What are those points?

c. I know the derivative is undefined at x=3 and x=1 (the x intercepts) but I am not sure why...
Again, look at the graph. The derivative is the "slope" of the tangent line" isn't it? What would a tangent line look like at x= 1 and x= 3? What do tangent lines look like at points just on either side of those?

d. Not sure where to go here, all i know is that i can put it in the form f(x)=|(x-3)(x-1)|, let u=x-3, let v=x-1, and that |u|=sqrt(u2), and |v|=sqrt(v2)
But I am not sure where to go from there..
At x= 3/2, you are well away from either x= 1 or x= 3. For all x close to 3/2 (x= 3/2+ h for some small h), f(x) is either $x^2- 4x+ 3$ or $-(x^2- 4x+ 3)$ Which is it? Can you differentiate that?

## 1. What is the definition of an absolute quadratic?

An absolute quadratic is a mathematical expression that contains a variable raised to the power of two, with no other terms or variables in the expression. It is also known as an absolute value quadratic because it is often used to represent a parabola that opens upwards or downwards.

## 2. How is an absolute quadratic different from a regular quadratic?

The main difference between an absolute quadratic and a regular quadratic is that an absolute quadratic has an absolute value sign around the variable, while a regular quadratic does not. This means that the absolute value of the variable must be taken before squaring it in an absolute quadratic, while a regular quadratic can have any value for the variable before squaring it.

## 3. What is the process for solving an absolute quadratic equation?

The process for solving an absolute quadratic equation involves two main steps: isolating the absolute value expression and then solving for the variable inside the absolute value. To isolate the absolute value expression, you must set the expression equal to a positive and negative version of itself. Then, solve each equation separately and check for extraneous solutions.

## 4. Can an absolute quadratic have more than one solution?

Yes, an absolute quadratic can have more than one solution. This is because the absolute value of a number can be positive or negative, so when solving an absolute quadratic equation, you will often get two solutions, one for each version of the equation (positive and negative).

## 5. How are absolute quadratic equations used in real life?

Absolute quadratic equations are used in real life to model various situations that involve a parabolic shape, such as the trajectory of a projectile, the shape of a suspension bridge, or the curve of a roller coaster. They are also used in optimization problems, where the goal is to find the maximum or minimum value of a function. Additionally, absolute quadratic equations are used in economics to model supply and demand curves.

Replies
27
Views
2K
Replies
5
Views
961
Replies
26
Views
1K
Replies
1
Views
785
Replies
7
Views
848
Replies
2
Views
608
Replies
3
Views
1K
Replies
8
Views
654
Replies
4
Views
1K
Replies
14
Views
773