Parts of the graphs of the functions $$h(x) = x$$ and k(x) = x^3, ext{ for } x ext{ in } \\mathbb{R},$$ are shown in the diagram below - Leaving Cert Mathematics - Question 6 - 2018

To find the area, we determine the integral of the difference between the functions from $x = -1$ to $x = 1$:

$ext{Area} = 2 imes \int_{-1}^{1} (h(x) - k(x)) \, dx$

Here, we calculate:

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

This simplifies to:

$= 2 \left[ \frac{x^2}{2} - \frac{x^4}{4} \right]_{-1}^{1}$

Evaluating this gives:

$= 2 \left( \left[ \frac{1^2}{2} - \frac{1^4}{4} \right] - \left[ \frac{(-1)^2}{2} - \frac{(-1)^4}{4} \right] \right)$

Which results in:

$= 2 \left( [\frac{1}{2} - \frac{1}{4}] - [\frac{1}{2} - \frac{1}{4}]\right) = 1 \, \text{unit}^2$.

To sketch the inverse function $k^{-1}(x)$ for the function defined by $k(x) = x^3$, we note that the inverse is given by:

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

The graph will be a reflection of the original function across the line $y = x$. The key points to plot would be:

ightarrow (0, 0)$$

ightarrow (1, 1)$$

ightarrow (-1, -1)$$

The graph of $k^{-1}(x)$ can be sketched by connecting these points and their symmetric counterparts.