Find the values of b and c that make f continuous everywhere.
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Steven G
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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.
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
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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
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\(f(x)=\begin{cases}\dfrac{x^2-6x+b}{x^2-7x+12} &: x>4 \\ 2\exp(x^2-3x+c) &: x\le 4\end{cases}\)
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 \(\large{\mathop {\lim }\limits_{x \to {4^ - }} f(x) = 2~?}\)
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
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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.
\(\displaystyle x^2- 6x+ 8= (x- 4)(x- 2)\) 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.
pka
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Do we have a new Soroban? He could never let a student discover something for themselves (a know-it-all).