Value to make function continuous

[lastochka]

Hello,
I have this exercise that I can't solve:

when x<3 the function f is given by the formula
f(x)=$\frac{4{x}^{3}-12{x}^{2}+10x-30}{x-3}$

when 3 < or =x
f(x)=$3{x}^{2}$-2x+a

What value must be chosen for a in order to make this function continuous at 3?

I think that I will have to equate both functions (may be I am wrong?), but before I have to replace with 3 for x... for the first function denominator will be 0... so I am not sure how to do it.
Please, can someone help!
Thank you!

 

[MarkFL]

We are given the piecewise function:

$$f(x)=\begin{cases}\dfrac{4x^3-12x^2+10x-30}{x-3}, & x<3 \\[3pt] 3x^2-2x+a, & 3\le x \\ \end{cases}$$

We first need to compute:

$$L=\lim_{x\to3^{-}}\frac{4x^3-12x^2+10x-30}{x-3}$$

Now, we find that the numerator is zero for $x=3$, and so we know $x-3$ is a factor, and the singularity is removable. Using synthetic division, we find:

$$\begin{array}{c|rr}& 4 & -12 & 10 & -30 \\ 3 & & 12 & 0 & 30 \\ \hline & 4 & 0 & 10 & 0 \end{array}$$

Thus we now know:

$$L=\lim_{x\to3^{-}}4x^2+10=46$$

Thus we require:

$$\lim_{x\to3^{+}}3x^2-2x+a=46$$

Can you proceed?

 

[lastochka]

Thank you MarkFL!
The answer is a=25
Thank you again!