Question

question 6 (1 point)
what is true about the function ( f(x)=-\frac{1}{x^{2}+6x - 7} ), as ( x
ightarrow1^{-} )?
( \bigcirc ) a) ( f(x)
ightarrow0 ) from below
( \bigcirc ) b) ( f(x)
ightarrow0 ) from above
( \bigcirc ) c) ( f(x)
ightarrowinfty )
( \bigcirc ) d) ( f(x)
ightarrow-infty )

Explanation:

Step1: Factor the denominator

Factor \(x^{2}+6x - 7=(x + 7)(x - 1)\). So \(f(x)=-\frac{1}{(x + 7)(x - 1)}\).

Step2: Analyze the limit as \(x\to1^{-}\)

When \(x\to1^{-}\), \(x-1\to0^{-}\) and \(x + 7\to8\). Then \((x + 7)(x - 1)\to0^{-}\). So \(\frac{1}{(x + 7)(x - 1)}\to-\infty\).

Step3: Find the limit of \(f(x)\)

Since \(f(x)=-\frac{1}{(x + 7)(x - 1)}\), then \(f(x)\to\infty\) as \(x\to1^{-}\).

Answer:

C. \(f(x)\to\infty\)