The function f has domain −2 ≤ x < 6 and is linear from (−2, 10) to (2, 0) and from (2, 0) to (6, 4) - Edexcel - A-Level Maths Pure - Question 1 - 2013 - Paper 7

To find f(0), we first locate 0 on the x-axis. It lies between the intervals (−2, 10) and (2, 0). Since f is linear between these points, we can use the equation of the line. The slope of the line from (−2, 10) to (2, 0) is:

$ext{slope} = \frac{0 - 10}{2 - (-2)} = \frac{-10}{4} = -2.5$

Using the point-slope form of a line:

$y - y_1 = m(x - x_1)$

Taking the point (−2, 10):

$y - 10 = -2.5(x + 2)$

At x = 0:

$y - 10 = -2.5(0 + 2)$ $y - 10 = -5$ $y = 5$

Thus, f(0) = 5.

To find the inverse of the function g, we set:

$g(x) = \frac{4 + 3x}{5 - x}$

We replace g(x) with y:

$y = \frac{4 + 3x}{5 - x}$

Now, we solve for x:

1. Multiply both sides by (5 - x): $y(5 - x) = 4 + 3x$
2. Expand: $5y - yx = 4 + 3x$
3. Rearrange to isolate x: $yx + 3x = 5y - 4$
4. Factor out x: $x(y + 3) = 5y - 4$
5. Thus: $x = \frac{5y - 4}{y + 3}$

Therefore, the inverse function is:

$g^{-1}(y) = \frac{5y - 4}{y + 3}$

To solve the equation, we first express g(f(x)):

We already know that:

$g(f(x)) = \frac{4 + 3f(x)}{5 - f(x)}$

Setting this equal to 16 gives:

$\frac{4 + 3f(x)}{5 - f(x)} = 16$

1. Cross-multiply: $4 + 3f(x) = 16(5 - f(x))$
2. Expand: $4 + 3f(x) = 80 - 16f(x)$
3. Rearranging gives: $3f(x) + 16f(x) = 80 - 4$
4. Combining terms yields: $19f(x) = 76$
5. Solving for f(x): $f(x) = \frac{76}{19} = 4$

Thus, the solution to the equation is:

$f(x) = 4$