Given f(x) = \(\log \frac{{{\rm{x}} + 4}}{{{\rm{x}} + 1}}\) and g(x) = \(2{\rm{x}} + {{\rm{x}}^2}\), then what is f[g(x)] equal to?

Question

Question

Answer 1

Concept:

Composition of function: fog(x) = f(g(x)), it means that the value of x for the function f(x) is g(x).

Example: Suppose f(x ) = x + 2 and g(x) = 2x. Then f(g(x)) = g(x) + 2 = 2x + 2

\(\log {{\bf{a}}^{\bf{n}}} = {\bf{n}}\log {\bf{a}}\)

Calculation:

Given that, f(x) = \(\log \frac{{{\rm{x}} + 4}}{{{\rm{x}} + 1}}\) and g(x) = \(2{\rm{x}} + {{\rm{x}}^2}\)

f[g(x)] = \(\log \frac{{{\rm{g}}\left( {\rm{x}} \right) + 4}}{{{\rm{g}}\left( {\rm{x}} \right) + 1}}\)

\( = \log \left( {\frac{{2{\rm{x}} + {{\rm{x}}^2} + 4{\rm{\;}}}}{{2{\rm{x}} + {{\rm{x}}^2} + 1}}} \right)\)

Hence, option (4) is correct.

Mistake Points
For option C, the Numerator should be (4x + x2 + 4) but it is given as (2x + x2 + 4). Therefore, option 4 is correct answer.