Sets, Relation and Function
117174
Let \(f: R \rightarrow R, g: R \rightarrow R\) be two functions given by \(f(x)=2 x-3, g(x)=x^3+5\). Then, \((f \circ g)^{-1}(x)\) is equal to
1 \(\left(\frac{x+7}{2}\right)^{\frac{1}{3}}\)
2 \(\left(x-\frac{7}{2}\right)^{\frac{1}{3}}\)
3 \(\left(\frac{x-2}{7}\right)^{\frac{1}{3}}\)
4 \(\left(\frac{\mathrm{x}-7}{2}\right)^{\frac{1}{3}}\)
Explanation:
Exp: (D) : Given,
\(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R} \text {, }\)
And,
\(\mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}\) be two function
Then,
\(\mathrm{f}(\mathrm{x})=2 \mathrm{x}-3, \quad \mathrm{~g}(\mathrm{x})=\mathrm{x}^3+5\)
\(\text { fog }(x)=f\{g(x)\}\)
\(\text { fog }(x)=f\left\{x^3+5\right\}\)
\(\text { fog }(x)=2\left(x^3+5\right)-3\)
\(\text { fog }(x)=2 x^3+10-3\)
\(\text { fog }(x)=2 x^3+7\)
Let,
\(f \circ g(x)=y\)
Then,
\(y=2 x^3+7\)
\(y-7=2 x^3\)
\(x^3=\frac{y-7}{2}\)
\(x=\left(\frac{y-7}{2}\right)^{1 / 3}=(f \circ g)^{-1}(y)\)So, \((f \circ g)^{-1}(x)=\left(\frac{x-7}{2}\right)^{1 / 3}\).
