Question

B. Determine the values of a and b so that f is continuous.

Please help!

Answer 1

Answer:

following are the solution to this question:

Step-by-step explanation:

Please find the complete question in the attached file.

$\lim_{x\to 2}+f(x) = \lim_{x\to 2}+ (ax^2-bx+3) = 4a-2b+3\\\\ \to 4a-2b+3 = 4\\\\ \therefore \ \ 4a-2b = 1$

The one-sided limits of F(x) at x = 3 must be equivalent for f(x) to be continuous at x = 3.

$\to \lim_{x\to 3}- f(x) = \lim_{x\to 3}- (ax^2-bx+3) = 9a-3b+3\\\\ \to \lim_{x\to 3}+ f(x) = \lim_{x\to 3}+ (2x-a+b) = 6-a+b$  

So,

$\to 9a-3b+3 = 6-a+b\\\\\therefore\\ \to 10a -4b = 3$

$\to 4a - 2b = 1......(a)\\\to 10a - 4b = 3.......(b)$

In equation a multiply the by -2 and then add in the equation b:

$\to -8a + 4b = -2\\\to 10a - 4b = 3\\ \to 2a = 1\\\\ \to a = \frac{1}{2}\\\\ \to \ 4(\frac{1}{2}) - 2b = 1\\\\ \to 2 - 2b = 1\\\\ \to -2b = -1\\\\ \to b = \frac{1}{2}$  

So, the value of $a \ and \ b= \frac{1}{2}$