Function ƒ(x): Continuity & Differentiability

[unf0r5ak3n]

Homework Statement

Let f be the function defined as ƒ(x)={ lx-1l + 2, for x<1, and ax^2 + bx, for x (greater or equal to) 1, where a and b are constants.

Homework Equations

A) If a=2 and b=3, is f continuous for all of x?
B) Describe all the values of a and b for which f is a continuous function?
C) For what values of a and b is f both continuous and differentiable?

 

[micromass]

It is clear that we only need to question if the function is continuous in 1.

You have probably seen that a function f is continuous in 1 iff

$$\lim_{x\rightarrow 1}{f(x)}=f(1)$$

Now, it is also true that a limit exists iff the right-hand and left-hand limits exists and are equal. So you only need to check that

$$\lim_{x\rightarrow 1+}{f(x)}=\lim_{x\rightarrow 1-}{f(x)}=f(1)$$

So calculate the left-hand limits and right-limits and see if they are equal to f(1)

 

[unf0r5ak3n]

so would the lim ƒ(x) = 5?

 

[micromass]

What is the limit

$$\lim_{x\rightarrow 1}{|x-1|+2}$$

and what is the limit

$$\lim_{x\rightarrow 1}{ax^2 + bx}$$

You want those limits to be equal in order for f to be continuous.

 

[unf0r5ak3n]

the limit of abs(x-1)+2 is 2 and the limit of ax^2+bx is 5 so the limit is not continuous right?

 

[micromass]

Oh, in question 1. Yes you are correct. f in question 1 is not continuous in 1.

 

[unf0r5ak3n]

for b) i said if a and b = 1 then lim of x^2+x as x approaches 1 is 2 which would make it a continuous function because the left and right limits are equal at f(1), is that correct? I am a bit confused with left and right hand limits

 

[micromass]

Yes, you are correct. But a=b=1 aren't the only possible values for a and b.

In general, we have that

$$\lim_{x\rightarrow 1}{ax^2+bx} =a+b$$

So every a and b such that a+b=2 suffices to make the function f continuous.

 

[unf0r5ak3n]

ah thank you i think I am understanding it now but I am confused about part c.

 

[micromass]

For part c, you will need to calculate the right and left limit of

$$\frac{f(x)-f(1)}{x-1}$$

They will need to be equal. This gives you another condition on a and b.