# Multiple solutions to f'(x)

1. Oct 26, 2008

1. The problem statement, all variables and given/known data

$$y=\frac{x^2}{1+x^2}$$ where $$-1\leq x \leq 1$$

The gradient at the point $$x=1$$ is $$\frac{1}{2}$$

Hence show that there is a point with $$\frac{1}{4} < x < \frac{1}{2}$$ where the gradient is also $$\frac{1}{2}$$

3. The attempt at a solution

I differentiated, set the derivative to $$\frac{1}{2}$$ and got to the point $$(x-1)(x^3 + x^2 + 3x - 1)=0$$, which is correct. Becoming stuck, I looked at the answers, and got more confused, the which says:

Let $$g(x)=x^3 + x^2 + 3x - 1$$, then

$$g(\frac{1}{4})=-\frac{11}{64} < 0$$
$$g(\frac{1}{2})=\frac{7}{8} > 0$$

Hence there is a root in the interval $$(\frac{1}{4},\frac{1}{2})$$

I don't understand how that shows that the root exists. Any help appreciated! :)

2. Oct 26, 2008

### CompuChip

Do you understand that the other point where the gradient is 1/2, is a root of g(x) ?
Because then it is just an application of the intermediate value theorem. Intuitively put: if you have a continuous graph which is below the axis at some point, and above it at some other point, then it must cross the axis somewhere in between (which makes sense, right? It cannot get from below to above without either crossing it, or jumping over it -- the latter is not allowed because it is continuous).

3. Oct 26, 2008