Question

1. algebraically, determine all invariant points, as an exact value, when $f(x)=\frac{1}{2}x^2 - 7$ is transformed to $y = \sqrt{f(x)}$. (2 marks)

Explanation:

Step1: Set up invariant point equation

Invariant points satisfy $f(x) = \sqrt{f(x)}$. Let $t = \sqrt{f(x)}$, so $t \geq 0$, and the equation becomes $t^2 = t$.

Step2: Solve for $t$

Rearrange: $t^2 - t = 0$
Factor: $t(t - 1) = 0$
Solutions: $t=0$ or $t=1$

Step3: Solve $f(x)=0$

Substitute $f(x)=\frac{1}{2}x^2 -7$:
$\frac{1}{2}x^2 -7 = 0$
$\frac{1}{2}x^2 =7$
$x^2=14$
$x=\pm\sqrt{14}$
Corresponding $y=0$, so points $(\sqrt{14},0)$ and $(-\sqrt{14},0)$.

Step4: Solve $f(x)=1$

Substitute $f(x)=\frac{1}{2}x^2 -7$:
$\frac{1}{2}x^2 -7 =1$
$\frac{1}{2}x^2=8$
$x^2=16$
$x=\pm4$
Corresponding $y=1$, so points $(4,1)$ and $(-4,1)$.

Step5: Verify domain validity

For $y=\sqrt{f(x)}$, $f(x)\geq0$. All solved $x$-values satisfy $\frac{1}{2}x^2 -7\geq0$, so all points are valid.

Answer:

The invariant points are $(-\sqrt{14}, 0)$, $(\sqrt{14}, 0)$, $(-4, 1)$, and $(4, 1)$.