Find a & b for g(x) Diff at x=1

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Homework Help Overview

The problem involves determining the values of a and b for the piecewise function g(x) defined as g(x) = 3x² for x≤1 and g(x) = a+bx for x>1, such that g is differentiable at x=1.

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  • Mixed

Approaches and Questions Raised

  • Participants discuss the continuity of g(x) at x=1 and the conditions for differentiability, including limits from both sides. There are questions about the correctness of the limits and the values of a and b based on the definitions of g(x).

Discussion Status

Some participants have offered guidance on the limits and the implications for the values of a and b, while others have questioned the assumptions made in the original poster's analysis. The discussion reflects a range of interpretations regarding the continuity and differentiability conditions.

Contextual Notes

There appears to be confusion regarding the continuity condition and the limits involved, particularly with respect to the values of a and b. The original poster's reference to limits and derivatives has prompted clarification and correction from other participants.

Janez25
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Homework Statement


For all real numbers a and b, define g(x) = 3x² if x≤1, a+bx if x>1. For what values of a and b is g differentiable at x=1?


Homework Equations





The Attempt at a Solution


g(x) is continuous: lim as x→1- [f(x)] = 1; lim as x→1+ [f(x)] = a+b
g(x) is differentiable: g'(x) = lim as x→1- of [(g(x)-g(1))/x-1] = 6 and g'(x) = lim as x→1+ [(g(x)-g(1))/x-1]=lim as x→1+ (b) = 0 {I think the limit from the right is correct}
I believe a=6, and I am not sure what b equals.
 
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Where did f come from? For the first part, lim as x -> 1- [g(x)] =/= 1, so a + b should equal a number that is not 1 in the resulting equation. Just as for your second analysis problem, think about what g'(x) should be for x < 1 and what it should be for x > 1.
 
Last edited:
Janez25 said:

Homework Statement


For all real numbers a and b, define g(x) = 3x² if x≤1, a+bx if x>1. For what values of a and b is g differentiable at x=1?


Homework Equations





The Attempt at a Solution


g(x) is continuous: lim as x→1- [f(x)] = 1; lim as x→1+ [f(x)] = a+b
g(x) is differentiable: g'(x) = lim as x→1- of [(g(x)-g(1))/x-1] = 6 and g'(x) = lim as x→1+ [(g(x)-g(1))/x-1]=lim as x→1+ (b) = 0 {I think the limit from the right is correct}
No, the limit from the right is NOT correct. For x> 1, g(x)= a+ bx and the derivative of a linear function is just its slope, b. Since the limit from the left is 6, you have b= 6, together with your previous equation, a+ b= 1.

I believe a=6, and I am not sure what b equals.
 
Ok, I have it now. Thanks!
 

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