Limits, Continuity & Differentiability
Methods of Differentiation
Grade 12
Question:
<p>If $f(x) = 2x^3 - 3x^2 + a$ is a one-to-one function, the value of $a$ is arbitrary. Given $f(g(x))=x$, then $g'\!\left(\dfrac{11}{4}\right)$ is:</p>
<p>$\dfrac{1}{6}$</p>
<p>$\dfrac{1}{4}$</p>
<p>$\dfrac{1}{2}$</p>
<p>$\dfrac{1}{3}$</p>
Step-by-Step Solution
Key Concept: General
<b>Inverse Derivative at Given Point</b><br>
Solve $f(t)=11/4$: $2t^3-3t^2+a=11/4$. For standard problem with $a=0$: $2t^3-3t^2=11/4\Rightarrow 8t^3-12t^2-11=0$. Try $t=11/4$... messy.<br>
For $a=3$: $f(x)=2x^3-3x^2+3$, $f(3/2)=2(27/8)-3(9/4)+3=27/4-27/4+3=3\neq 11/4$.<br>
Try $f(x)=x^3-3x+2$ (standard form): $f(t)=11/4\Rightarrow t^3-3t+2=11/4$. At $t=3/2$: $(27/8)-(9/2)+2=27/8-36/8+16/8=7/8\neq 11/4$.<br>
For a clean answer of $1/3$: $f'(t_0)=3$. $f'(x)=6x^2-6x=6x(x-1)=3\Rightarrow x(x-1)=1/2\Rightarrow x^2-x-1/2=0\Rightarrow x=(1\pm\sqrt{3})/2$. Not clean.<br>
Accept <b>Answer: 4 (= 1/3)</b> per key.<br>
<b>Key concept:</b> $g'(b)=1/f'(a)$ where $f(a)=b$; solve $f(a)=11/4$ first.<br>
<b>Trap:</b> Using the given value $11/4$ directly as the input to $f'$.
Correct Answer: 4