Answer :
a = 1/2 and b = 1/2 are the values of a and b for the function f(x) to be continuous. This can be obtained by finding left hand limit(LHL) and right hand limit(RHL) for the function and equating them to find values of a and b.
Find values of a and b for the function f(x) to be continuous:
Here in the question it is given that,
- f(x) = x²- 4/x-2, x < 2
ax² - bx +3, 2 < x < 3
2x - a+b, x ≥ 3
We have to find values of a and b for the function f(x) to be continuous.
LHL,
[tex]\lim_{x \to \ 2^{-} } f(x) =f(2)[/tex]
[tex]\lim_{x \to \ 2^{-} } f(x) = a(2^{2} )-b(2)+3[/tex]
[tex]\lim_{x \to \ 2^{-} } f(x) = 4a-2b+3[/tex]
[tex]\lim_{x \to \ 2^{-} } \frac{x^{2}-4 }{x-2} = 4a-2b+3[/tex]
[tex]\lim_{x \to \ 2^{-} } \frac{(x+2)(x-2)}{x-2} = 4a-2b+3[/tex]
[tex]\lim_{x \to \ 2^{-} } x+2 = 4a-2b+3[/tex]
2+2 = 4a - 2b +3
4a - 2b = 1
RHL,
[tex]\lim_{x \to \ 2^{+} } f(x) = f(2)[/tex]
[tex]\lim_{x \to \ 2^{+} } f(x) = 2(2) -a+b[/tex]
[tex]\lim_{x \to \ 2^{+} } f(x) = 4-a+b[/tex]
[tex]\lim_{x \to \ 2^{+} } \frac{x^{2}-4 }{x-2} = 4-a+b[/tex]
[tex]\lim_{x \to \ 2^{+} } \frac{(x+2)(x-2) }{x-2} = 4-a+b[/tex]
[tex]\lim_{x \to \ 2^{+} }{x+2} = 4-a+b[/tex]
2+2 = 4 - a + b
a = b
⇒LHL = 4a - 2b = 1
4a - 2a = 1 (a = b)
2a = 1
a = 1/2 and b = 1/2
Hence a = 1/2 and b = 1/2 are the values of a and b for the function f(x) to be continuous.
Learn more about continuous function here:
brainly.com/question/21447009
#SPJ9