Function

Algebra Level 2

If $f(x) = 2x^4 - {4}$ and $g(x) = \sqrt{\dfrac{x-2}{\sqrt{2}}}$ , then what is the value of $f \circ g(5)$ ?

5

2

\sqrt{2}

\sqrt{5}

2 solutions

Chew-Seong Cheong
Sep 15, 2017

$\begin{aligned} f \circ g (5) & = f(g({\color{#D61F06}5})) \\ & = f \left(\sqrt{\frac {{\color{#D61F06}5}-2}{\sqrt 2}} \right) \\ & = f \left({\color{#3D99F6}\sqrt{\frac 3{\sqrt 2}}} \right) \\ & = 2\left({\color{#3D99F6}\sqrt{\frac 3{\sqrt 2}}} \right)^4 - 4 \\ & = 2\left(\frac 92 \right) - 4 \\ & = 9 - 4 \\ & = \boxed{5} \end{aligned}$

@Sakib Nazmus , use f \circ g $f \circ g$

Chew-Seong Cheong - 3 years, 8 months ago

@Chew-Seong Cheong didn't understand the fourth line.

Nazmus sakib - 3 years, 8 months ago

Thanks. A typo. I have corrected it.

Chew-Seong Cheong - 3 years, 8 months ago

A Former Brilliant Member
Sep 25, 2017

$g(5)=\sqrt{\dfrac{5-2}{\sqrt{2}}}=\sqrt{\dfrac{3}{\sqrt{2}}}$

$f \circ g(5)=2\left(\sqrt{\dfrac{2}{\sqrt{2}}}\right)^4-4=2\left(\dfrac{3}{\sqrt{2}}\right)^2-4=2\left(\dfrac{9}{2}\right)-4=9-4=$ $\large{\color{#D61F06}\boxed{5}}$

You made a simple mistake. It should be $g(5)=\sqrt{\dfrac{5-2}{\sqrt{2}}}=\sqrt{\dfrac{3}{\sqrt{2}}}$ and $f \circ g(5)=2\left(\sqrt{\dfrac{3}{\sqrt{2}}}\right)^4-4$

Nazmus sakib - 3 years, 8 months ago

Function

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