Verify Rolle's theorem for the function $f(x)=x^2+5x-6$ in the interval (-6,1)
After checking continuous for the function. I saw the function is continuous and differentiable. After checking differentiability I got $f'(x)=2x+5$. I was following steps to verify Rolle's theorem.
$$f'(c)=0$$
$$2c+5=0$$
$$c=-\frac{5}{2}$$
I wait for a moment. Then, I looked at book. I saw they wrote that
$$c=-\frac{5}{2}\in (-6,1)$$
How $-\frac{5}{2}$ is element of -6 to 1? I think I am confusing here something.
Best Answer
Your question isn't very clear. If you are asking why $$c=-\frac{5}{2}\in (-6,1)$$
Then see $-\dfrac{5}{2}$ is bigger than $-6$ (why?) and $-\dfrac{5}{2}$ is smaller than $1$ (why?). That is why it is lying in this range