Given the functions f(x) = x2, g(x) = rac1x−4, and h(x) = kx+m. a) Explain the meaning of x eq 4 in g(x). b) Find g(0). c) Find g⁻¹(0). d) Find k and m if f ext o h(x) = 9x2 − 12x + 4.

Here are the solutions for questions 1 and 3.

Question 1:
Given the functions $f(x) = x^2$, $g(x) = \frac{1}{x-4}$, and $h(x) = kx+m$.

a) Explain the meaning of the statement "$x \neq 4$" in the function $g(x)$.
The statement $x \neq 4$ means that $x$ can be any real number except 4. If $x$ were equal to 4, the denominator of $g(x)$ would become zero, making the function undefined.

b) Find $g(0)$.

Step 1: Substitute $x=0$ into the function $g(x)$.

$$g(0) = \frac{1}{0-4}$$

Step 2: Simplify the expression.

$$g(0) = \frac{1}{-4} = -\frac{1}{4}$$

The value of $g(0)$ is $\boxed{-\frac{1}{4}}$.

c) Find $g^{-1}(0)$.

Step 1: Find the inverse function $g^{-1}(x)$.
Let $y = g(x) = \frac{1}{x-4}$.
Swap $x$ and $y$:

$$x = \frac{1}{y-4}$$

Step 2: Solve for $y$.

$$x(y-4) = 1$$ $$y-4 = \frac{1}{x}$$ $$y = \frac{1}{x} + 4$$

So, $g^{-1}(x) = \frac{1}{x} + 4$.

Step 3: Substitute $x=0$ into $g^{-1}(x)$.

$$g^{-1}(0) = \frac{1}{0} + 4$$

Division by zero is undefined.
The value of $g^{-1}(0)$ is $\boxed{\text{undefined}}$.

d) The value of $k$ and $m$ given that $f \circ h(x) = 9x^2 - 12x + 4$.

Step 1: Find the composite function $f \circ h(x)$.

$$f \circ h(x) = f(h(x)) = f(kx+m)$$

Substitute $h(x)$ into $f(x)$:

$$f(kx+m) = (kx+m)^2$$

Step 2: Expand the expression.

$$(kx+m)^2 = k^2x^2 + 2kmx + m^2$$

Step 3: Equate this to the given expression $9x^2 - 12x + 4$.

$$k^2x^2 + 2kmx + m^2 = 9x^2 - 12x + 4$$

Step 4: Compare the coefficients of $x^2$, $x$, and the constant terms.
From the coefficient of $x^2$:

$$k^2 = 9 \implies k = \pm 3$$

From the constant term:

$$m^2 = 4 \implies m = \pm 2$$

From the coefficient of $x$:

$$2km = -12$$

Step 5: Determine the correct pairs of $k$ and $m$.
If $k=3$:

$$2(3)m = -12$$ $$$$