Jambmaths question:
Find the value of x for which the function $f(x)=2{{x}^{3}}-{{x}^{2}}-4x+4$has a maximum value
Option A:
-1
Option B:
$-\frac{2}{3}$
Option C:
1
Option D:
$\frac{2}{3}$
Jamb Maths Solution:
$\begin{align} & f(x)=2{{x}^{3}}-{{x}^{2}}-4x+4 \\ & {{f}^{1}}(x)=6{{x}^{2}}-2x-4 \\ & \text{At stationary point }f(x)=0 \\ & 6{{x}^{2}}-2x-4=0 \\ & 3{{x}^{2}}-x-2=0 \\ & (3x+2)(x-1)=0 \\ & x=-\frac{2}{3}\text{ or }x=1 \\ & \text{To test for min or max} \\ & f''(x)=\frac{d}{dx}(6{{x}^{2}}-2x-4) \\ & {{f}^{''}}(x)=12x-2 \\ & \text{At }x=-\frac{2}{3} \\ & {{f}^{''}}(x)=12(-\tfrac{2}{3})-2=-10<0\text{ (max point)} \\ & \text{ At }x=-\tfrac{2}{3}\text{ will give a maximum value for the function} \\\end{align}$
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