Parts of the graphs of the functions $h(x) = x$ and $k(x) = x^3$, $x \in \mathbb{R}$, are shown in the diagram below. (a) Find the co-ordinates of the points of intersection of the graphs of the two functions. (b) (i) Find the total area enclosed between the graphs of the two functions. (ii) On the diagram on the previous page, using symmetry or otherwise, draw the graph of $k^{-1}$, the inverse function of $k$.

Leaving Cert Mathematics Graphs of Functions: Parts of the graphs of the funct

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

Question 6

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Parts of the graphs of the functions $h(x) = x$ and $k(x) = x^3$, $x \in \mathbb{R}$, are shown in the diagram below. (a) Find the co-ordinates of the points of... show full transcript

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

Step 1

Find the co-ordinates of the points of intersection of the graphs of the two functions.

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Answer

To find the points of intersection, we set the two functions equal to each other:

$x^3 = x$

Rearranging the equation yields:

$x^3 - x = 0$

Factoring out an $x$ gives:

$x(x^2 - 1) = 0$

This factors to:

$x(x - 1)(x + 1) = 0$

The solutions are:

$x = 0, \quad x = 1, \quad x = -1$

Next, we substitute back to find the corresponding $y$ -values:

For $x = 0: h(0) = 0, k(0) = 0 \Rightarrow (0, 0)$
For $x = 1: h(1) = 1, k(1) = 1 \Rightarrow (1, 1)$
For $x = -1: h(-1) = -1, k(-1) = -1 \Rightarrow (-1, -1)$

Thus, the points of intersection are: $(-1, -1)$ , $(0, 0)$ , and $(1, 1)$ .

Step 2

Find the total area enclosed between the graphs of the two functions.

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Answer

To find the total area between the graphs, we set up the integral:

$ext{Area} = 2 \int_0^1 (x - x^3) \, dx$

Calculating the integral:

$= 2 \left[ \frac{x^2}{2} - \frac{x^4}{4} \right]_0^1 = 2 \left[ \frac{1^2}{2} - \frac{1^4}{4} \right]$ $= 2 \left[ \frac{1}{2} - \frac{1}{4} \right]$ $= 2 \left[ \frac{1}{4} \right] = \frac{1}{2} \text{ unit}^2$

Step 3

On the diagram on the previous page, using symmetry or otherwise, draw the graph of $k^{-1}$, the inverse function of $k$.

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Answer

To draw the inverse function $k^{-1}(x)$ , we reflect the graph of $k(x) = x^3$ across the line $y = x$ .

Since $k^{-1}(x)$ is defined as $k^{-1}(x) = \sqrt[3]{x},$ we can plot the following points to aid in the graphing:

For $x = 0: k^{-1}(0) = 0$
For $x = 1: k^{-1}(1) = 1$
For $x = -1: k^{-1}(-1) = -1$

The final graph should show a curve that passes through these points, illustrating the reflection of the cubic function.

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

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

Find the co-ordinates of the points of intersection of the graphs of the two functions.

Find the total area enclosed between the graphs of the two functions.

On the diagram on the previous page, using symmetry or otherwise, draw the graph of $k^{-1}$, the inverse function of $k$.

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