# Show that the equation x^4 + 2x^2 − 2 = 0 has exactly one solution on [0, 1]?

Jun 25, 2018

See below.

#### Explanation:

First of all, let's compute $f \left(x\right) = {x}^{4} + 2 {x}^{2} - 2$ at the boundary of our domain:

$f \left(0\right) = {0}^{4} + 2 \cdot {0}^{2} - 2 = - 2 < 0$

$f \left(1\right) = {1}^{4} + 2 \cdot {1}^{2} - 2 = 1 > 0$

If we compute the derivative

$f ' \left(x\right) = 4 {x}^{3} + 4 x = 4 x \left({x}^{2} + 1\right)$

We can see that it is always positive in $\left[0 , 1\right]$. In fact, ${x}^{2} + 1$ is always positive, and $4 x$ is obviously positive, since $x$ is positive.

So, our function starts below the $x$ axis, since $f \left(0\right) < 0$, and ends above the $x$ axis, since $f \left(1\right) > 0$. The function is a polynomial, and so it is continuous.

If a continuous line starts below the axis and ends above, it means that it must have crossed it somewhere in between. And the fact that the derivative is always positive means that the function is always growing, and so it can't cross the axis twice, hence the proof.