The graph of $f(x) = a^x$, where $a > 0$ and $a \neq 1$, passes through the point \(\left( 3, \frac{27}{8} \right)\) - NSC Mathematics - Question 5 - 2016 - Paper 1

The inverse function equation can be derived as follows:

Starting from ( y = a^x ), we take logs:

$x = log_a(y).$

Rearranging gives:

$y = a^x \text{ or expressed as } x = log_a(y) \Rightarrow f^{-1}(y) = log_a(y).$

To find ( f'^{-1}(x) ) we first need ( f'(x) ). Since ( f(x) = a^x ), we know that:

$f'(x) = a^x \ln(a).$

Setting this equal to -1, we evaluate:

( f'(x) = -1 ) leads to:

$a^x \ln(a) = -1.$

There may not be valid real solutions based on the range restrictions of the functions.

The domain of $h(x) = f(x - 5)$ will be influenced by the function's domain. Since $f(x) = a^x$ is defined for all real numbers, $h(x)$ is also defined for all real numbers. Therefore, the domain of $h$ is:

$\text{Domain of } h: (-\infty, \infty).$

In sketching the graph of (g(x) = a b^x + q) for $a < 0$, $b > 1$, and $q < 0$:

• The graph will decrease and will have a horizontal asymptote at $y = q$.
• The $x$-intercept can be found by solving ( 0 = a b^x + q ) for $x$ and will lie below the $y$-axis.
• It will not cross the $x$-axis if $q < 0$.
• The $y$-intercept happens when $x = 0$: ( g(0) = ab^0 + q = a + q ).