Differentiating absolute quadratic

[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..

 

[Mark44]

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.

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..

 

[HallsofIvy]

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?