question 6 (1 point)\nwhat is true about the function ( f(x)=-\frac{1}{x^{2}+6x - 7} ), as (…

question 6 (1 point)\nwhat is true about the function ( f(x)=-\frac{1}{x^{2}+6x - 7} ), as ( x\rightarrow1^{-} )?\n( \bigcirc ) a) ( f(x)\rightarrow0 ) from below\n( \bigcirc ) b) ( f(x)\rightarrow0 ) from above\n( \bigcirc ) c) ( f(x)\rightarrowinfty )\n( \bigcirc ) d) ( f(x)\rightarrow-infty )

question 6 (1 point)\nwhat is true about the function ( f(x)=-\frac{1}{x^{2}+6x - 7} ), as ( x\rightarrow1^{-} )?\n( \bigcirc ) a) ( f(x)\rightarrow0 ) from below\n( \bigcirc ) b) ( f(x)\rightarrow0 ) from above\n( \bigcirc ) c) ( f(x)\rightarrowinfty )\n( \bigcirc ) d) ( f(x)\rightarrow-infty )

Answer

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)