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) Find the total area enclosed between the graphs of the two functions. (i) On the diagram on the previous page, using symmetry or otherwise, draw the graph of $k^{-1}(x)$, the inverse function of $k$.

Leaving Cert Mathematics Integration: 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 i... 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 equations equal to each other:

$x = x^3$

Rearranging gives us:

$x^3 - x = 0$

Factoring, we have:

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

This yields:

$x = 0 \\ \\ x^2 - 1 = 0 \\ \\ x = 1 \\text{ or }\\ x = -1$

Thus, the points of intersection are:

When $x = -1$ , $y = -1 \Rightarrow (-1, -1)$
When $x = 0$ , $y = 0 \Rightarrow (0, 0)$
When $x = 1$ , $y = 1 \Rightarrow (1, 1)$

The co-ordinates of 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 area between the graphs, we need to evaluate the integral from $x = -1$ to $x = 1$ of the difference between the two functions:

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

First, we simplify the integral:

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

Calculating this gives:

$2 \left[ \frac{x^2}{2} - \frac{x^4}{4} \right]_{0}^{1} = 2 \left(\frac{1^2}{2} - \frac{1^4}{4} - (0 - 0)\right) = 2 \left(\frac{1}{2} - \frac{1}{4}\right) = 2 \cdot \frac{1}{4} = \frac{1}{2}$

Thus, the total area enclosed between the graphs is: $\frac{1}{2} \text{ unit}^2$ .

Step 3

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

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Answer

To sketch the graph of the inverse function $k^{-1}(x)$ , we note that $k(x) = x^3$ implies $k^{-1}(x) = \sqrt[3]{x}$ .

The key points to plot are:

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

Since the graph of $k^{-1}(x)$ is obtained by reflecting the graph of $k(x)$ across the line $y = x$ , the shape will be mirrored accordingly.

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 $k^{-1}(x)$, the inverse function of $k$.

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