Let \( f(x)=\sqrt{42-x} \) and \( g(x)=x^{2}-x \) Then the domain of \( f \circ g \) is equal to Question Help: Video

To find the domain of the composite function \( f \circ g \), we need to consider the composition of functions \( f \) and \( g \). Given: \( f(x) = \sqrt{42-x} \) \( g(x) = x^{2} - x \) The composition of functions \( f \circ g \) is denoted as \( f(g(x)) \). Substitute \( g(x) \) into \( f(x) \): \( f(g(x)) = f(x^{2} - x) = \sqrt{42 - (x^{2} - x)} \) Now, we need to find the domain of \( f(g(x)) \), which is the set of all real numbers for which the expression is defined. The expression \( \sqrt{42 - (x^{2} - x)} \) is defined when the expression inside the square root is non-negative, i.e., \( 42 - (x^{2} - x) \geq 0 \). Solve the inequality \( 42 - (x^{2} - x) \geq 0 \) to find the domain of \( f \circ g \). Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(42-\left(x^{2}-x\right)\geq 0\) - step1: Remove the parentheses: \(42-x^{2}+x\geq 0\) - step2: Rewrite the expression: \(42-x^{2}+x=0\) - step3: Factor the expression: \(\left(7-x\right)\left(6+x\right)=0\) - step4: Separate into possible cases: \(\begin{align}&7-x=0\\&6+x=0\end{align}\) - step5: Solve the equation: \(\begin{align}&x=7\\&x=-6\end{align}\) - step6: Determine the test intervals: \(\begin{align}&x<-6\\&-67\end{align}\) - step7: Choose a value: \(\begin{align}&x_{1}=-7\\&x_{2}=1\\&x_{3}=8\end{align}\) - step8: Test the chosen value: \(\begin{align}&x<-6\textrm{ }\textrm{is not a solution}\\&-67\textrm{ }\textrm{is not a solution}\end{align}\) - step9: Include the critical value: \(\begin{align}&-6\leq x\leq 7\textrm{ }\textrm{is the solution}\end{align}\) - step10: The final solution is \(-6\leq x\leq 7:\) \(-6\leq x\leq 7\) The solution to the inequality \( 42 - (x^{2} - x) \geq 0 \) is \( -6 \leq x \leq 7 \). Therefore, the domain of the composite function \( f \circ g \) is \( -6 \leq x \leq 7 \).