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 8 - 2013 - Paper 7

To find f(0), we need to identify which segment of the piecewise function applies. For x = 0, it lies in the segment between (−2, 10) and (2, 0). The linear equation for this segment can be derived using the points (−2, 10) and (2, 0).

The slope (m) is:

$m = \frac{0 - 10}{2 - (-2)} = -\frac{10}{4} = -\frac{5}{2}$

Using point-slope form:

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

Substituting in one of the points, say (−2, 10):

$y - 10 = -\frac{5}{2}(x + 2)$

Simplifying gives:

$y = -\frac{5}{2}x - 5 + 10$

So,

$f(x) = -\frac{5}{2}x + 5$

Then for f(0):

$f(0) = -\frac{5}{2}(0) + 5 = 5$

To find the inverse of the function g defined as:

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

We start by setting y = g(x):

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

Cross-multiplying gives:

$y(5 - x) = 4 + 3x$

Expanding and rearranging:

$5y - yx = 4 + 3x$

Grouping terms involving x:

$yx + 3x = 5y - 4$

Factor out x:

$x(y + 3) = 5y - 4$

Solving for x, we get:

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

Thus, the inverse function is:

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

We need to find x such that:

$g(f(x)) = 16$

First, substitute f(x) into g:

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

Set it equal to 16:

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

Cross-multiplying the equation:

$4 + 3f(x) = 16(5 - f(x))$

Expanding gives:

$4 + 3f(x) = 80 - 16f(x)$

Grouping all terms involving f(x):

$3f(x) + 16f(x) = 80 - 4$

$19f(x) = 76$

Solving for f(x):

$f(x) = \frac{76}{19} = 4$

Now we have f(x) = 4. We can find the corresponding x value:

From our earlier piecewise definition, we see f(x) = 4 at x = 6.

Thus, the solution to the equation g(f(x)) = 16 is:

$x = 6$