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For which parameter values the function is continuous and differentiable

Yankel

Active member
Jan 27, 2012
398
For which values of a,b and c, the next function is continuous and differentiable at x=2 ?

[tex]\left\{\begin{matrix} 3x-1 & x\leq 2\\ ax^{2}+bx+c & x>2 \end{matrix}\right.[/tex]

1. b=2-c
2. b=6+2c+2a
3. 7+c-2a
4. b=3-a-(3/4)c

I know that f(2)=5, and so is the limit of f when x goes to 2 from the left side.
I have calculated the limit when f goes to 2 from the right side, and I got:

4a+2b+c

and so for continuously I need:

4a+2b+c = 5

but here I got stuck...
 

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
The only point you need to look at is $x = 2$, as all the others must be continuous and differentiable (remember all polynomials are well-behaved).

Recall that for a function to be continuous at $a$, you need:

$$\lim_{x \to a^{-}} f(x) = \lim_{x \to a^{+}} f(x)$$

So in your case, you need to show that the following is true:

$$\lim_{x \to 2^{-}} 3x - 1 = \lim_{x \to 2^{+}} ax^2 + bx + c ~ ~ ~ \Longleftrightarrow ~ ~ ~ 5 = 4a + 2b + c$$

Which is what you got, so this is correct. But, you also need differentiability.

Recall that for a function $f(x)$ to be differentiable at $a$, you need:

$$\lim_{x \to a^{-}} f'(x) = \lim_{x \to a^{+}} f'(x)$$

So, differentiate both pieces of the function, and show that the following is true:

$$\lim_{x \to 2^{-}} 3 = \lim_{x \to 2^{+}} 2ax + b ~ ~ ~ \Longleftrightarrow ~ ~ ~ 3 = 4a + b$$

You are now left with two equations in three unknowns:

$$4a + 2b + c = 5$$

$$4a + b = 3$$

Subtract the second from the first, to obtain:

$$b + c = 2 ~ ~ ~ \Longleftrightarrow b = 2 - c$$

And substitute this back into the second:

$$4a + 2 - c = 3 ~ ~ ~ \Longleftrightarrow ~ ~ ~ a = \frac{1}{4} \left ( c + 1 \right )$$

This gives you $a$ and $b$ in terms of $c$. Choose any $c$, for instance $c = 2$, and you get:

$$a = \frac{3}{4} ~ ~ ~ \text{and} ~ ~ ~ b = 0$$

That is one solution, out of infinitely many. This is also the only class of solutions.
 
Last edited:

chisigma

Well-known member
Feb 13, 2012
1,704
For which values of a,b and c, the next function is continuous and differentiable at x=2 ?

[tex]\left\{\begin{matrix} 3x-1 & x\leq 2\\ ax^{2}+bx+c & x>2 \end{matrix}\right.[/tex]

1. b=2-c
2. b=6+2c+2a
3. 7+c-2a
4. b=3-a-(3/4)c

I know that f(2)=5, and so is the limit of f when x goes to 2 from the left side.
I have calculated the limit when f goes to 2 from the right side, and I got:

4a+2b+c

and so for continuously I need:

4a+2b+c = 5

but here I got stuck...
You have to impose first that $\displaystyle \lim_{x \rightarrow 2 +} f(x) = \lim_{x \rightarrow 2 -} f(x)$ and after that $\displaystyle \lim_{x \rightarrow 2 +} f ^{\ '} (x) = \lim_{x \rightarrow 2 -} f^{\ '}(x)$...


Kind regards

$\chi$ $\sigma$
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
The only point you need to look at is $x = 2$, as all the others must be continuous and differentiable (remember all polynomials are well-behaved).

Recall that for a function to be continuous at $a$, you need:

$$\lim_{x \to a^{-}} f(x) = \lim_{x \to a^{+}} f(x)$$

So in your case, you need to show that the following is true:

$$\lim_{x \to 2^{-}} 3x - 1 = \lim_{x \to 2^{+}} ax^2 + bx + c ~ ~ ~ \Longleftrightarrow ~ ~ ~ 5 = 4a + 2b + c$$

Which is what you got, so this is correct. But, you also need differentiability.

Recall that for a function $f(x)$ to be differentiable at $a$, you need:

$$\lim_{x \to a^{-}} f'(x) = \lim_{x \to a^{+}} f'(x)$$
This is true, but not an obvious statement. It looks like the definition of "continuous" but derivatives are NOT necessarily contuinuous. What is true is that any derivative, even if not continuous, satisfies the "intermediate value theorem".

So, differentiate both pieces of the function, and show that the following is true:

$$\lim_{x \to 2^{-}} 3 = \lim_{x \to 2^{+}} 2ax + b ~ ~ ~ \Longleftrightarrow ~ ~ ~ 3 = 4a + b$$

You are now left with two equations in three unknowns:

$$4a + 2b + c = 5$$

$$4a + b = 3$$

Subtract the second from the first, to obtain:

$$b + c = 2 ~ ~ ~ \Longleftrightarrow b = 2 - c$$

And substitute this back into the second:

$$4a + 2 - c = 3 ~ ~ ~ \Longleftrightarrow ~ ~ ~ a = \frac{1}{4} \left ( c + 1 \right )$$

This gives you $a$ and $b$ in terms of $c$. Choose any $c$, for instance $c = 2$, and you get:

$$a = \frac{3}{4} ~ ~ ~ \text{and} ~ ~ ~ b = 0$$

That is one solution, out of infinitely many. This is also the only class of solutions.