Find the co-ordinates of the points of intersection of the graphs of the two functions - Leaving Cert Mathematics - Question 6 - 2018

To find the area between the curves $h(x) = x$ and $k(x) = x^3$, we compute the integral of the top function minus the bottom function between the points of intersection:

$ext{Area} = 2 \int_{0}^{1} (x - x^3) \, dx$

First, we evaluate the integral:

$\int (x - x^3) \, dx = \frac{x^2}{2} - \frac{x^4}{4}$

Now, applying the limits from 0 to 1:

$ext{Area} = 2 \left[ \frac{1^2}{2} - \frac{1^4}{4} \right] - 2 \left[ \frac{0^2}{2} - \frac{0^4}{4} \right]$

Calculating this gives:

$= 2 \left[ \frac{1}{2} - \frac{1}{4} \right] = 2 \left[ \frac{1}{4} \right] = \frac{1}{2} \text{ units}^2$

To graph the inverse function $k^{-1}(x)$ of $k(x) = x^3$, we swap the $x$ and $y$ coordinates. The inverse function is given by:

$k^{-1}(x) = \sqrt[3]{x}$

This is a reflection of $k(x)$ across the line $y = x$. The graph will pass through points such as:

• $(0, 0)$,
• $(1, 1)$,
• $(-1, -1)$

The sketch should show this relation, indicating that wherever $k(x)$ rises steeply in positive $x$, $k^{-1}(x)$ will also rise steeply in positive $y$.