Given: $f(x) = x(x - 3)^2$ with $f'(1) = f'(3) = 0$ and $f(1) = 4$. 8.1 Show that $f$ has a point of inflection at $x = 2$. 8.2 Sketch the graph of $f$, clearly indicating the intercepts with the axes and the turning points. 8.3 For which values of $x$ will $y = -f(x)$ be concave down? 8.4 Use your graph to answer the following questions: 8.4.1 Determine the coordinates of the local maximum of $h$ if $h(x) = f(x - 2) + 3$. 8.4.2 Claire claims that $f''(2) = 1$. Do you agree with Claire? Justify your answer.

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Given: $f(x) = x(x - 3)^2$ with $f'(1) = f'(3) = 0$ and $f(1) = 4$ - NSC Mathematics - Question 8 - 2017 - Paper 1

Question 8

Given: $f(x) = x(x - 3)^2$ with $f'(1) = f'(3) = 0$ and $f(1) = 4$. 8.1 Show that $f$ has a point of inflection at $x = 2$. 8.2 Sketch the graph of $f$, clearly ... show full transcript

Worked Solution & Example Answer:Given: $f(x) = x(x - 3)^2$ with $f'(1) = f'(3) = 0$ and $f(1) = 4$ - NSC Mathematics - Question 8 - 2017 - Paper 1

Step 1

Show that $f$ has a point of inflection at $x = 2$

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Answer

Step 1: Find the first and second derivatives.

First derivative is given by:

$f'(x) = 3x^2 - 6x + 9$

To find the second derivative:

$f''(x) = 6x - 6$

Step 2: Set the second derivative to zero.

Solving for $f''(x) = 0$ :

$6x - 6 = 0 \Rightarrow x = 1$

Step 3: Determine the sign change.

Evaluating around $x = 2$ :

For $x < 2$ , $f''(1) = 0$ , hence $f$ is concave down.
For $x > 2$ , $f''(3) = 6$ , so $f$ is concave up.

Thus, there is a point of inflection at $x = 2$ .

Step 2

Sketch the graph of $f$, clearly indicating the intercepts with the axes and the turning points.

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Answer

Step 4: Finding intercepts.

To find the y-intercept, set $x = 0$ : $f(0) = 0(0 - 3)^2 = 0$

To find x-intercepts, set $f(x) = 0$ : $x(x - 3)^2 = 0 \Rightarrow x = 0 \text{ or } x = 3$

Step 5: Finding turning points.

We already found that $f'(1) = 0$ behaves as a maximum since the derivative changes signs from positive to negative.

The turning points are $(1, 4)$ and a horizontal tangent at $(3, 0)$ .

Step 6: Graph sketching.

Draw the graph indicating the intercepts $(0,0)$ , $(3,0)$ , and turning point $(1,4)$ .

Step 3

For which values of $x$ will $y = -f(x)$ be concave down?

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Answer

The function $y = -f(x)$ will be concave down where $f'(x)$ is decreasing. Since we established that $f''(x) < 0$ for $x < 2$ , then:

$y = -f(x) \text{ is concave down for } x > 2.$

Step 4

Determine the coordinates of the local maximum of $h$ if $h(x) = f(x - 2) + 3$.

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Answer

To find the local maximum of $h$ , we identify that $h'(x) = f'(x-2)$ , which leads us to find:

Setting $h'(x) = 0$ yields: $f'(x - 2) = 0 \Rightarrow (x - 2) = 1 \Rightarrow x = 3.$

Plugging $x = 3$ back into $h$ : $h(3) = f(1) + 3 = 4 + 3 = 7.$ The coordinates of the local maximum of $h$ are $(3, 7)$ .

Step 5

Do you agree with Claire? Justify your answer.

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Answer

No, I do not agree with Claire.

To check if $f''(2) = 1$ : $f''(2) = 6(2) - 6 = 6.$ Since $f''(2) = 6$ , Claire’s statement is incorrect.

Given: $f(x) = x(x - 3)^2$ with $f'(1) = f'(3) = 0$ and $f(1) = 4$ - NSC Mathematics - Question 8 - 2017 - Paper 1

Worked Solution & Example Answer:Given: $f(x) = x(x - 3)^2$ with $f'(1) = f'(3) = 0$ and $f(1) = 4$ - NSC Mathematics - Question 8 - 2017 - Paper 1

Show that $f$ has a point of inflection at $x = 2$

Sketch the graph of $f$, clearly indicating the intercepts with the axes and the turning points.

For which values of $x$ will $y = -f(x)$ be concave down?

Determine the coordinates of the local maximum of $h$ if $h(x) = f(x - 2) + 3$.

Do you agree with Claire? Justify your answer.

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