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This is one of our many quizzes on Evaluating functions. Click on this to access more such competitive quizzes on Functions.

Evaluating functions Quiz 2

mathematics

 If, $f\left( x \right) = 10x - 5$ and $g\left( x \right) = 5x + 1$. Find $\left( {f + g} \right)\left( x \right)$.

 Substitute the given value of $f(x)$ and $g(x)$ in $\left( {f + g} \right)\left( x \right) = f\left( x \right) + g\left( x \right)$.

 Given, $f\left( x \right) = 10x - 5$ and $g\left( x \right) = 5x + 1$ $ \\ $ W.K.T $\left( {f + g} \right)\left( x \right) = f\left( x \right) + g\left( x \right)$ $ \\ $ On substituting $ \\ \left( {f + g} \right)\left( x \right) = \left( {10x - 5} \right) + \left( {5x + 1} \right) \\ \Rightarrow \left( {f + g} \right)\left( x \right) = 15x - 4$

 Given, $f\left( x \right) = ax + 3$ and $g\left( x \right) = 3x$, for what value of $a$ is $\left( {f \circ g} \right)\left( 4 \right) = 9$

 Compose the function and evaluate for $4$.

 Given, $f\left( x \right) = ax + 3$ and $g\left( x \right) = 3x$ $ \\ $ Now, compose the function and evaluate for $4$: $\left( {f \circ g} \right)\left( 4 \right) = f\left( {g\left( 4 \right)} \right) = f\left( {3\left( 4 \right)} \right) = a\left( {3\left( 4 \right)} \right) + 3 \\ \left( {f \circ g} \right)\left( 4 \right) = f\left( {g\left( 4 \right)} \right) = 12a + 3 \\ $ Now equate to the given, $\left( {f \circ g} \right)\left( 4 \right) = 9$: $ \\ \Rightarrow 12a + 3 = 9 \\ \Rightarrow 12a = 6 \\ \Rightarrow a = \dfrac{1}{2}$

 Let $g\left( x \right) = {x^2} - 5x + 1$, then evaluate for $g\left( {2x - 1} \right)$.

 Substitute the value of the given function and apply ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ to evaluate.

 Given, $g\left( x \right) = {x^2} - 5x + 1$ $ \\ $ Substitute the value of the given function i.e., $g\left( {2x - 1} \right)$, we get: $ \\ g\left( {2x - 1} \right) = {\left( {2x - 1} \right)^2} - 5\left( {2x - 1} \right) + 1 \\ \Rightarrow g\left( {2x - 1} \right) = \left( {{{\left( {2x} \right)}^2} - 4x + 1} \right) - 5\left( {2x - 1} \right) + 1 \\ \Rightarrow g\left( {2x - 1} \right) = 4{x^2} - 4x + 1 - 10x + 5 + 1 \\ \Rightarrow g\left( {2x - 1} \right) = 4{x^2} - 14x + 7$

 The function $g\left( x \right) = \dfrac{1}{2}x + 2$ , find composite of a given function $f\left( x \right) =  - 10{x^2} + 6x - 4$

 Substitute the value of the given input and then evaluate the terms.

 Given, $f\left( x \right) =  - 10{x^2} + 6x - 4$ and $g\left( x \right) = \dfrac{1}{2}x + 2$ $ \\ $ Substitute the value of the given input as: $ \\ g\left[ {f\left( x \right)} \right] = \dfrac{1}{2}\left[ {f\left( x \right)} \right] + 2 \\ \Rightarrow g\left[ {f\left( x \right)} \right] = \dfrac{1}{2}\left[ { - 10{x^2} + 6x - 4} \right] + 2 \\ \Rightarrow g\left[ {f\left( x \right)} \right] =  - 5{x^2} + 3x - 2 + 2 \\ \Rightarrow g\left[ {f\left( x \right)} \right] =  - 5{x^2} + 3x$

 Simplify $f\left( x \right) = \dfrac{{\dfrac{x}{{x + 2}} - \dfrac{3}{5}}}{{x - 3}}$ and apply limits to the simplified expression at $x \to 5$.

 To simplify the given expression, simplify based on its power value of the given exponent.

 Given, $f\left( x \right) = \dfrac{{\dfrac{x}{{x + 2}} - \dfrac{3}{5}}}{{x - 3}}$ $ \\ $ Multiply and divide the function by $5\left( {x + 2} \right)$ $ \\ f\left( x \right) = \dfrac{{\dfrac{x}{{x + 2}} - \dfrac{3}{5}}}{{x - 3}} \cdot \dfrac{{5\left( {x + 2} \right)}}{{5\left( {x + 2} \right)}} \\ \Rightarrow f\left( x \right) = \dfrac{{5x - 3\left( {x + 2} \right)}}{{5\left( {x + 2} \right)\left( {x - 3} \right)}} \\ \Rightarrow f\left( x \right) = \dfrac{{2x - 6}}{{5\left( {x + 2} \right)\left( {x - 3} \right)}} \\ \Rightarrow f\left( x \right) = \dfrac{{2\left( {x - 3} \right)}}{{5\left( {x + 2} \right)\left( {x - 3} \right)}} \\ \Rightarrow f\left( x \right) = \dfrac{2}{{5\left( {x + 2} \right)}} \\ $ Now, applying limits to the obtained expression at $x \to 5$ i.e., $ \\ f\left( x \right) = \mathop {\lim }\limits_{x \to 5} \dfrac{2}{{5\left( {x + 2} \right)}} \\ \Rightarrow f\left( x \right) = \dfrac{2}{{5\left( 7 \right)}} \\ \Rightarrow f\left( x \right) = \dfrac{2}{{35}}$