| Als f (x) = x² + 2 en |
&space;\right&space;)=\frac{x^2+2}{\sqrt{x^2+4}}} "\mathbf{g\: \left (f(x) \right )=\frac{x^2+2}{\sqrt{x^2+4}}}") |
| dan is g (x) | | gelijk aan |
|
A. x² + 2 |
|---|
| B. x² + 4 |
| C. \(\frac{x}{\sqrt{x+2}}\) |
| D. \(\frac{x}{\sqrt{x^2+2}}\) |
| E. \(\frac{x^2+2}{\sqrt{x^2+4}}\) |
[ 5-8583 - op net sinds 8.2.2021-(E)-19.10.2026 ]
Translation in E N G L I S H
| If f (x) = x² + 2 and |
&space;\right&space;)=\frac{x^2+2}{\sqrt{x^2+4}}} "\mathbf{g\: \left (f(x) \right )=\frac{x^2+2}{\sqrt{x^2+4}}}") |
| then g (x) | | equals |
|
A. x² + 2 |
|---|
| B. x² + 4 |
| C. \(\frac{x}{\sqrt{x+2}}\) |
| D. \(\frac{x}{\sqrt{x^2+2}}\) |
| E. \(\frac{x^2+2}{\sqrt{x^2+4}}\) |
Oplossing - Solution
We lopen de alternatieven af :
Als g(x) = x² + 2 dan is g(f(x)) = (x² + 2)² + 2 → NIET A
Als g(x) = x² + 4 dan is g(f(x)) = (x² + 2)² + 4 → NIET B
\(Als\; g(x) = \frac{x}{\sqrt{x+2}}\; dan \; is \; g(f(x)) = \frac{x^2+2}{\sqrt{x^2+2+2}} =\frac{x^2+2}{\sqrt{x^2+4}}\)
