# Question #88f84

Jul 20, 2016

We need to calculate both sides of the equation, and find a value $c$ that satisfies it

#### Explanation:

We calculate first $f \left(2\right) = 2 \cdot {2}^{2} - 3 \cdot 2 + 1 = 3$
We then calculate $f \left(0\right) = 2 \cdot {0}^{2} - 3 \cdot 0 + 1 = 1$

We then calculate $\frac{f \left(2\right) - f \left(0\right)}{2 - 0} = \frac{3 - 1}{2 - 0} = 1$

We calculate now $f ' \left(x\right) = 4 x - 3$, and now we need to find a point $c$ where $f ' \left(c\right) = 1$, that is $4 c - 3 = 1$, but this means $4 c = 4$, and hence $c = 1$ is the value we are looking for