(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

[tex]y=\frac{x^2}{1+x^2}[/tex] where [tex]-1\leq x \leq 1[/tex]

The gradient at the point [tex]x=1[/tex] is [tex]\frac{1}{2}[/tex]

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

3. The attempt at a solution

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

Let [tex]g(x)=x^3 + x^2 + 3x - 1[/tex], then

[tex]g(\frac{1}{4})=-\frac{11}{64} < 0[/tex]

[tex]g(\frac{1}{2})=\frac{7}{8} > 0[/tex]

Hence there is a root in the interval [tex](\frac{1}{4},\frac{1}{2})[/tex]

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

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# Multiple solutions to f'(x)

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