Verify Rolle's theorem

Question

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.

Answer 1

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

Answer 2

The fraction $-\frac{5}{2}$ is equal to the decimal $-2.5$, which is clearly between $-6$ and $1$.