Homework problem

[cal quw wus69]

Find the values of a and b that make f continuous everywhere. (Enter your answers as fractions.)


x^2-1/x-1 if x<1
f(x)= ax^2-bx+1 if 1<=x<3
4x-a+b if x >=3


a=
b=

 

[Deleted member 4993]

Find the values of a and b that make f continuous everywhere. (Enter your answers as fractions.)


x^2-1/x-1 if x<1
f(x)= ax^2-bx+1 if 1<=x<3
4x-a+b if x >=3


a=
b=

What are the conditions of continuity of a function?

Please share your work with us, indicating exactly where you are stuck so that we may know where to begin to help you.

 

[galactus]

This is what you have written:

\(\displaystyle f(x)=\left\{\begin{array}{rcl} x^{2}-\frac{1}{x}-1, \;\ \mbox{if} \;\ x<1\\ax^{2}-bx+1, \;\ \mbox{if} \;\ \;\ 1\leq x<3\\4x-a+b, \;\ \text{if} \;\ \;\ x\geq 3\end{array}\right\)

For the first one, I assume you mean $\displaystyle \frac{x^{2}-1}{x-1}$?. If so, please use proper grouping symbols.

 

[cal quw wus69]

ahh yeah i do, sorry bout that

 

[BigGlenntheHeavy]

$\displaystyle galactus, \ that's \ 4x-a+b, \ not \ 4a-x+b \ if \ x \ \ge \ 3.$

 

[galactus]

Thanks for the catch. A typo.

 

[BigGlenntheHeavy]

\(\displaystyle f(x)=\left\{\begin{array}{rcl} x+1, \;\ \mbox{if} \;\ x<1\\ax^{2}-bx+1, \;\ \mbox{if} \;\ \;\ 1\leq x<3\\4x-a+b, \;\ \text{if} \;\ \;\ x\geq 3\end{array}\right\)

$\displaystyle See \ graph \ and \ see \ if \ you \ can \ figured \ it \ out, \ not \ that \ hard.$

$\displaystyle Hint: \ My \ way, \ one \ has \ to \ cheat \ a \ little.$

$\displaystyle For \ f(x) \ to \ be \ continuous \ throughout, \ x+1 \ and \ ax^2-bx+1 \ must \ meet \ at \ x=1$

$\displaystyle and \ ax^2-bx+1 \ and \ 4x-a+b \ must \ meet \ at \ x \ = \ 3.$

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