consider the following function.\n f(x)=\frac{x^{2}-17 x + 72}{x - 8} \n(a) explain why ( f ) has a…

consider the following function.\n f(x)=\frac{x^{2}-17 x + 72}{x - 8} \n(a) explain why ( f ) has a removable discontinuity at ( x = 8 ). (select all that apply.)\n( lim _{x \rightarrow 4} f(x) ) does not exist.\n( f(8) ) is undefined.\n( lim _{x \rightarrow 8} f(x) ) is finite.\n( f(8) ) and ( lim _{x \rightarrow 8} f(x) ) are finite, but are not equal.\nnone of the above\n(b) &edefine ( f(8) ) so that ( f ) is continuous at ( x = 8 ) (and thus the discontinuity is removed).\n( f(8)= )
Answer
Explanation:
Step1: Factor the numerator
We factor (x^{2}-17x + 72) as ((x - 8)(x-9)). So (f(x)=\frac{(x - 8)(x - 9)}{x - 8}), for (x\neq8).
Step2: Simplify the function (for (x\neq8))
Cancel out the common factor ((x - 8)) (since (x\neq8)), we get (f(x)=x - 9) for (x\neq8).
Step3: Analyze the limit as (x\rightarrow8)
(\lim_{x\rightarrow8}f(x)=\lim_{x\rightarrow8}(x - 9)). Substitute (x = 8) into (x-9), we have (\lim_{x\rightarrow8}f(x)=8 - 9=-1). And (f(8)) is undefined because when (x = 8), the original function (f(x)=\frac{x^{2}-17x + 72}{x - 8}) has a denominator of (0).
For part (a):
- (\lim_{x\rightarrow4}f(x)) is not relevant to the discontinuity at (x = 8).
- Since (f(x)=\frac{x^{2}-17x + 72}{x - 8}), when (x = 8), the denominator is (0), so (f(8)) is undefined.
- We found (\lim_{x\rightarrow8}f(x)=-1) (finite).
- Since (f(8)) is undefined (not finite), the statement “(f(8)) and (\lim_{x\rightarrow8}f(x)) are finite, but are not equal” is wrong.
For part (b): To make (f(x)) continuous at (x = 8), we re - define (f(8)) to be equal to (\lim_{x\rightarrow8}f(x)). Since (\lim_{x\rightarrow8}f(x)=-1), we set (f(8)=-1).
Answer:
(a)
- (f(8)) is undefined.
- (\lim_{x\rightarrow8}f(x)) is finite. (b) (f(8)=-1)