By the mean value theorem, c is such that
[tex]\displaystyle f(c) = \frac1{2-(-2)} \int_{-2}^2 f(x) \, dx[/tex]
[tex]\displaystyle 3c(c-2) = \frac34 \int_{-2}^2 (x^2-2x) \, dx[/tex]
To quickly compute the integral, we have
[tex]\displaystyle \int_{-2}^2 (x^2-2x) \, dx = \int_{-2}^2 x^2 \, dx - 2 \int_{-2}^2 x \, dx[/tex]
but x is an odd function, and the integral of an odd function over a symmetric interval is zero; meanwhile x² is even, and the integral of an even function over a symmetric interval is twice the integral over half that interval:
[tex]\displaystyle \int_{-2}^2 (x^2-2x) \, dx = 2 \int_0^2 x^2 \, dx + 0[/tex]
[tex]\displaystyle \int_{-2}^2 (x^2-2x) \, dx = \frac23 x^3 \bigg|_0^2[/tex]
[tex]\displaystyle \int_{-2}^2 (x^2-2x) \, dx = \frac23 (2^3-0^3) = \frac{16}3[/tex]
Now solve for c :
[tex]3c^2 - 6c = \dfrac34 \times \dfrac{16}3[/tex]
[tex]3c^2 - 6c - 4 = 0[/tex]
[tex]\implies c = \dfrac{3 \pm \sqrt{21}}3[/tex]
but only one of these is in the range [-2, 2],
[tex]c = \dfrac{3 - \sqrt{21}}3 \approx -0.527525 \approx \boxed{-0.528}[/tex]
