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=(1)/(4)\\ x=-(1)/(2) \vee x=(1)/(2)\\ x\in\left\{-(1)/(2),0,(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\{-(1)/(2),(1)/(2)\right\}`,
`y''=0` because the factor
`(4x^2-1)^2=0`.
You have to check values of
`y'` around
`x\in\left\{-(1)/(2),(1)/(2)\right\}`.


`\forall x\in(\infty,-(1)/(2))\, y'<0 \Rightarrow y\searrow\\ \forall x\in(-(1)/(2),0)\, y'>0 \Rightarrow y\\earrow\\\\ \forall x\in(0,(1)/(2))\, y'<0 \Rightarrow y\searrow\\ \forall x\in((1)/(2),\infty)\, y'>0 \Rightarrow y\\earrow`

The above means that at
`x=-(1)/(2)` and
`x=-(1)/(2)` there are (global) maxima.