Suppose \( f(x)=-8 x+2 \) and \( g(x)=\frac{1}{x^{2}-2 x+6} \) \( (f \circ g)(x)=\square \)

To find \( (f \circ g)(x) \), we need to substitute \( g(x) \) into \( f(x) \). Starting with \( g(x) \): \[ g(x) = \frac{1}{x^2 - 2x + 6} \] Now substituting \( g(x) \) into \( f(x) \): \[ f(g(x)) = f\left(\frac{1}{x^2 - 2x + 6}\right) \] Using the function \( f(x) = -8x + 2 \): \[ f\left(\frac{1}{x^2 - 2x + 6}\right) = -8\left(\frac{1}{x^2 - 2x + 6}\right) + 2 \] This simplifies to: \[ = -\frac{8}{x^2 - 2x + 6} + 2 \] Thus, the final result for \( (f \circ g)(x) \) is: \[ (f \circ g)(x) = -\frac{8}{x^2 - 2x + 6} + 2 \]