Question

consider the following function.
f(x)=\frac{x^{2}-17 x + 72}{x - 8}
(a) explain why ( f ) has a removable discontinuity at ( x = 8 ). (select all that apply.)
( lim _{x
ightarrow 4} f(x) ) does not exist.
( f(8) ) is undefined.
( lim _{x
ightarrow 8} f(x) ) is finite.
( f(8) ) and ( lim _{x
ightarrow 8} f(x) ) are finite, but are not equal.
none of the above
(b) &edefine ( f(8) ) so that ( f ) is continuous at ( x = 8 ) (and thus the discontinuity is removed).
( f(8)= )

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 \(x
eq8\).

Step2: Simplify the function (for \(x

eq8\))
Cancel out the common factor \((x - 8)\) (since \(x
eq8\)), we get \(f(x)=x - 9\) for \(x
eq8\).

Step3: Analyze the limit as \(x

ightarrow8\)
\(\lim_{x
ightarrow8}f(x)=\lim_{x
ightarrow8}(x - 9)\). Substitute \(x = 8\) into \(x-9\), we have \(\lim_{x
ightarrow8}f(x)=8 - 9=-1\). 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

ightarrow4}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

ightarrow8}f(x)=-1\) (finite).

• Since \(f(8)\) is undefined (not finite), the statement “\(f(8)\) and \(\lim_{x

ightarrow8}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 \(\lim_{x
ightarrow8}f(x)\). Since \(\lim_{x
ightarrow8}f(x)=-1\), we set \(f(8)=-1\).

Answer:

(a)

• \(f(8)\) is undefined.
• \(\lim_{x

ightarrow8}f(x)\) is finite.
(b) \(f(8)=-1\)