Obtain the maximum value of the function \(f(x)=x^{3}-12 x+4\)
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Correct Answer: Option A
Explanation:
\(f( x )= x ^{3}-12 x +4\). At turning point, \(d / dx \quad f( x )=0\)
\begin{aligned}
\Rightarrow & 3 x ^{2}-12=0 \\
& X ^{2}=4 \\
\Rightarrow & x =\pm \sqrt{ } 4=\pm 2 \\
\text { At } x =2, & f(2)=2^{3}-12(2)+4 \\
=& 8-24+4=-12 \\
\text { At } x =-2, f(-2)=(-2)^{3}-12(-2)+4 \\
=&-8+24+4=20
\end{aligned}
Hence, the maximum value of the function is 20
\(f( x )= x ^{3}-12 x +4\). At turning point, \(d / dx \quad f( x )=0\)
\begin{aligned}
\Rightarrow & 3 x ^{2}-12=0 \\
& X ^{2}=4 \\
\Rightarrow & x =\pm \sqrt{ } 4=\pm 2 \\
\text { At } x =2, & f(2)=2^{3}-12(2)+4 \\
=& 8-24+4=-12 \\
\text { At } x =-2, f(-2)=(-2)^{3}-12(-2)+4 \\
=&-8+24+4=20
\end{aligned}
Hence, the maximum value of the function is 20