\(y=(4x^2-1)^4\\
y'=4(4x^2-1)^3\cdot8x=32x(4x^2-1)^3\\\\
32x(4x^2-1)^3=0\\
x=0 \vee 4x^2-1=0\\
4x^2-1=0\\
4x^2=1\\
x^2=\dfrac{1}{4}\\
x=-\dfrac{1}{2} \vee x=\dfrac{1}{2}\\
x\in\left\{-\dfrac{1}{2},0,\dfrac{1}{2}\right\}\leftarrow \text{stationary points}\\\\\)
\(y''=32((4x^2-1)^3+x\cdot3(4x^2-1)^2\cdot8x)\\
y''=32(4x^2-1)^2(4x^2-1+24x^2)\\
y''=32(4x^2-1)^2(28x^2-1)\\\\
y''(0)=32(4\cdot0^2-1)^2(28\cdot0^2-1)=32\cdot(-1)^2\cdot(-1)=-32\)
\(-32<0 \Rightarrow\) there's a (local) maximum at \(x=0\).
For \(x\in\left\{-\dfrac{1}{2},\dfrac{1}{2}\right\}\), \(y''=0\) because the factor \((4x^2-1)^2=0\).
You have to check values of \(y'\) around \(x\in\left\{-\dfrac{1}{2},\dfrac{1}{2}\right\}\).
\(\forall x\in(\infty,-\frac{1}{2})\, y'<0 \Rightarrow y\searrow\\
\forall x\in(-\frac{1}{2},0)\, y'>0 \Rightarrow y
earrow\\\\
\forall x\in(0,\frac{1}{2})\, y'<0 \Rightarrow y\searrow\\
\forall x\in(\frac{1}{2},\infty)\, y'>0 \Rightarrow y
earrow\)
The above means that at \(x=-\dfrac{1}{2}\) and \(x=-\dfrac{1}{2}\) there are (global) maxima.
