A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
Determine the values of a and b for which the following function is continuous at every real number x
f(x) { ax+b if x ≤ 1
2bx^2  8a if 1 < x < 2
4b^2/a 5x if x ≥ 2 }
anonymous
 3 years ago
Determine the values of a and b for which the following function is continuous at every real number x f(x) { ax+b if x ≤ 1 2bx^2  8a if 1 < x < 2 4b^2/a 5x if x ≥ 2 }

This Question is Open

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0continuous implies \[\lim_{x \rightarrow c}f(x)=f(c)\] for the first one, \[\lim_{x \rightarrow 1}f(x)=f(1)\] \[\lim_{x \rightarrow 1}(ax+b)=a+b\] and so on...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[f(x)=\begin{cases}ax+b&\text{if }x\le1\\ 2bx^28a&\text{if }1<x<2\\ \frac{4b^2}{a5x}&\text{if }x\ge2\end{cases}\] For f(x) to be continuous at x = 1 and x = 2, the following must be satisfied: \[\color{red}{\lim_{x\to1^}f(x)=\lim_{x\to1^+}f(x)}\\ \color{blue}{\lim_{x\to2^}f(x)=\lim_{x\to2^+}f(x)}\] \[RED:\lim_{x\to1^}(ax+b)=\lim_{x\to1^+}(2bx^28a)\\ BLUE: \lim_{x\to2^}(2bx^28a)=\lim_{x\to2^+}\frac{4b^2}{a5x}\] Determining the limits should give you a system of equations (assuming I rewrote the function you provided).
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.