Find the values of b and c that make f continuous everywhere.

[laslo]

Hello,
I need to find b and c parameters for this function, but I can't handle the e number.


Thank you in advance for your help.

 

[Steven G]

You say that you can't handle the e number meaning that you tried. Great, so can we see your work so we know where you need help. Thanks.

 

[laslo]

 

[Steven G]

You can't solve it because you do not know what it should equal! Besides you really should know what 16-12 equals.

You want the 2nd limit to exist, not be undefined. So think about what you want x^2 - 6x + b to equal when x=4. Then make that happen by defining b appropriately.

Please post back.

 

[HallsofIvy]

If the denominator if a fraction goes to 0 but the denominator does NOT then the limit of the fraction does not exist! What does that tell you about what $\displaystyle x^2- 6x+ b$ must be when x= 4 and what b must be?

 

[pka]

$f(x)=\begin{cases}\dfrac{x^2-6x+b}{x^2-7x+12} &: x>4 \\ 2\exp(x^2-3x+c) &: x\le 4\end{cases}$
If $b=8$ then using l'Hopital rule $\mathop {\lim }\limits_{x \to {4^ + }} f(x) = 2$.
Can you find a value of $c$ such that $\large{\mathop {\lim }\limits_{x \to {4^ - }} f(x) = 2~?}$

 

[HallsofIvy]

I wouldn't think you need to use "L'Hopital". The only reason you need to look close is because both numerator and denominator are 0 when x= 4. And, since these are polynomials, that means that x-4 is a factor of both.

$\displaystyle x^2- 6x+ 8= (x- 4)(x- 2)$ and
$\displaystyle x^2- 7x+ 12= (x- 4)(x- 3)$ so

$\displaystyle \lim_{x\to 4}\frac{x^2- 6x+ 8}{x^2- 7x+ 12}= \lim_{x\to 4}\frac{x- 2}{x- 3}= \frac{4- 2}{4- 3}= 2$.

Then, in order to have continuity at x= 4 we must have $\displaystyle \lim_{x\to 4} 2e^{x^2- 3x+ c}= 2$
$\displaystyle \lim_{x\to 4} 2e^{x^2- 3x+ c}= 2e^{16- 6+ c}= 2e^{10+ c}= 2$.

So we must have $\displaystyle e^{10+ c}= 1$ which means we must have 10+ c= 1 or c= -9.

 

[Steven G]

You have <7 minutes to fix your error. The powers should be 16-12+c = 4+c

 

[pka]

You have <7 minutes to fix your error. The powers should be 16-12+c = 4+c

Do we have a new Soroban? He could never let a student discover something for themselves (a know-it-all).

 

[laslo]

Thank you for your help. Now I see what was my mistake.