# How do I show that something only has one stationary point?

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.

• By the way, using the binomial expansion was unnecessary. Use the chain rune to get $dy/dx=8(1+2x)^3-8(1-2x)^3$, then you're looking for $x$ such that $(1+2x)^3=(1-2x)^3$. Real numbers only ever have one real cube root, so you're now trying to solve $1+2x=1-2x$, giving $x=0$. – user329501 Apr 9 '16 at 18:19
• @ProbablyWrong The question had marks based on the binomial expansion - but I'll bear that in mind for next time! – Soup Apr 9 '16 at 18:21

Notice that for $x\neq 0$, $\ x^2$ is always positive!
Notice that $3>0$ and that $x^2\geq 0 \forall x\in \mathbb{R}$. So there is no way that adding these two terms can ever be less than $3$(nevermind equalling $0$). I.e there will be no real roots of $3+4x^2$
• No problem. $\forall$ means "for all" and $x\in \mathbb{R}$ just means that $x$ is a real number. So now you know. – K.Power Apr 9 '16 at 18:16
The equation $3+4x^2=0$ does not have solution in the domain of real numbers because we have $x^2\geq0$ so $4x^2\geq0$ so $3+4x^2>4x^2\geq0$ so it follows $3+4x^2>0$