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

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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.
 
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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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