Question

Ex 2.6 9. Find the stationary points on the curve y= (4x²-1)^4 and determine their nature

Answer 1

$y=(4x^2-1)^4\\ y'=4(4x^2-1)^3\cdot8x=32x(4x^2-1)^3\\\\ 32x(4x^2-1)^3=0\\ x=0 \vee 4x^2-1=0\\ 4x^2-1=0\\ 4x^2=1\\ x^2=\dfrac{1}{4}\\ x=-\dfrac{1}{2} \vee x=\dfrac{1}{2}\\ x\in\left\{-\dfrac{1}{2},0,\dfrac{1}{2}\right\}\leftarrow \text{stationary points}$

$y''=32((4x^2-1)^3+x\cdot3(4x^2-1)^2\cdot8x)\\ y''=32(4x^2-1)^2(4x^2-1+24x^2)\\ y''=32(4x^2-1)^2(28x^2-1)\\\\ y''(0)=32(4\cdot0^2-1)^2(28\cdot0^2-1)=32\cdot(-1)^2\cdot(-1)=-32$
$-32$ 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 $x\in\left\{-\dfrac{1}{2},\dfrac{1}{2}\right\}$.

$\forall x\in(\infty,-\frac{1}{2})\, y'$

The above means that at $x=-\dfrac{1}{2}$ and $x=-\dfrac{1}{2}$ there are (global) maxima.