# how to find the values of?

Discussion in 'Linear and Abstract Algebra' started by remy, Nov 12, 2021.

1. ### remy

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how to i find what a,b or c by myself?

the question is what their values if the function is
Continuous

thanks

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remy, Nov 12, 2021

2. ### HallsofIvy

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If that is all you are given, a, b, and c can be anything at all. Is it possible that you were asked to find a, b, and c such that f is continuous?

If so then the limit as x approaches the "break point" must be equal to the value at that point or the limit from the other side.

For example, in the first problem you are given that f(x)= (x^3+ ax^2+ bx+ c)/(x+1)^2} for all x except at x= 1 and that f(-1)= -3. We can try to find the limit, as x goes to -1, by setting x= -1 but that makes the denominator 0. In order that the limit exist at all, and we want it to be -3, the numerator must also be 0 at x= -1. In fact, it must have a factor of (x+ 1)^2 to cancel the (x+ 1)^2 in the denominator.

Since the denominator is third degree it must be of the form (x+ 1)^2(px+ q) for some numbers a and b. That is (x^2+ 2x+ 1)(px+ q)= px^3+ 2px^2+ px+ qx^2+ 2qx+ q= px^3+ (2p+ q)x^2+ (p+ 2q)x+ q. That must be equal to x^3+ ax^2+ bx+ c. In order that two polynomials be the same for all x, they must have the same coefficients so we must have p= 1, 2p+ q= q+ 2= a, so q= a- 2, p+ 2q= 1+ 2a- 4=2a- 3= b. and q= a- 2= c. That is, px+ q= x+ a- 2 and when x= -1, -1+ a- 2= a- 3= -3. So a= 0, b= 2a- 3= -3, and c= a- 2= -2.

Check: (x+ 1)^2(x- 2)= (x^2+ 2x+ 1)(x- 2)= x^3- 2x^2+ 2x^2- 4x+ x- 2= x^3- 3x- 2
f(x)= (x^3- 3x- 2)/(x+ 1)^2= ((x+1)^2(x- 2))/(x+1)^2= x- 2 as long as x is not -1. The limit as x goes to -1 is -1- 2= -3, f(-1) so f is continuous at x= -1.

Last edited: Nov 17, 2021
HallsofIvy, Nov 17, 2021
remy likes this.

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