Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
alyannahere
Group Title
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 }
 one year ago
 one year ago
alyannahere Group Title
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 }
 one year ago
 one year ago

This Question is Open

sriramkumar Group TitleBest ResponseYou've already chosen the best response.0
continuous 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...
 one year ago

SithsAndGiggles Group TitleBest 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).
 one year ago
See more questions >>>
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.