I have this question in my homework where I can't quite grasp the solution.
I have been asked to show that a curve with equation $y=(1+2x)^4+(1-2x)^4$ has one singular stationary point and what its coordinates are. Simplifying using binomial expansion, I get $y=2+48x^2+32x^4$ and differentiating that for the gradient function I get $96x+128x^3$.
I know that to have a stationary point, the gradient must be zero so I put $96x+128x^3=0$. I then factorised it to get $32x(3+4x^2)=0$
Now's where the trouble I'm having comes in. I can see that something must be zero in order for the equation to equal zero. If $32x=0$ then $x=0$ at the stationary point. Why can't I say $3+4x^2=0$? The mark scheme I have says $3+4x^2>0$ but I can't see how they reached this conclusion.
Many thanks for your help in advance.
Best Answer
Notice that for $x\neq 0$, $\ x^2$ is always positive!