Limits, Continuity & Differentiability
Methods of Differentiation
Grade 12

Question:

<p>Suppose that $f(0) = 0$ and $f'(0) = 2$. Let $g(x) = f(-f(-f(x)))$. The value of $g'(0)$ is:</p>
<p>$1$</p>
<p>$0$</p>
<p>$8$</p>
<p>$-8$</p>

Step-by-Step Solution

Key Concept: General
<b>Chain Rule with Negation</b><br> $g(x) = f(-f(-f(x)))$.<br> Differentiating: $g'(x) = f'(-f(-f(x)))\cdot(-1)\cdot f'(-f(x))\cdot(-1)\cdot f'(x)$.<br> $= f'(-f(-f(x)))\cdot f'(-f(x))\cdot f'(x)$.<br> At $x=0$: $f(0)=0$, so $-f(0)=0$, $f(-f(0))=f(0)=0$, $-f(0)=0$.<br> $g'(0) = f'(0)\cdot f'(0)\cdot f'(0) = 2^3 = 8$.<br> <b>Key concept:</b> The two negation signs cancel: $(-1)\cdot(-1)=1$, so all three $f'$ factors are evaluated at 0.<br> <b>Trap:</b> Forgetting that $-f(0)=0$ (since $f(0)=0$), and mistakenly getting a sign error.
Correct Answer: C

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