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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 $x = 2$ - NSC Mathematics - Question 8 - 2017 - Paper 1

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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-$x-=-2$-NSC Mathematics-Question 8-2017-Paper 1.png

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 indic... show full transcript

Worked Solution & Example Answer: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 $x = 2$ - NSC Mathematics - Question 8 - 2017 - Paper 1

Step 1

Do you agree with Claire? Justify your answer.

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Answer

I do not agree with Claire; her statement is incorrect. To evaluate f(2)f'(2), we need to calculate it as follows:

f(2)=3(2)212(2)+9=1224+9=3f'(2) = 3(2)^2 - 12(2) + 9 = 12 - 24 + 9 = -3

This indicates that f(2)=3f'(2)=-3, not 11. Thus, Claire's assertion that f(2)=1f'(2) = 1 is false.

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