[tex]f'(x)\ge0[/tex] for all [tex]x[/tex] in [-3, 0], so [tex]f(x)[/tex] is non-decreasing over this interval, and in particular we know right away that its minimum value must occur at [tex]x=-3[/tex].
From the plot, it's clear that on [-3, 0] we have [tex]f'(x)=-x[/tex]. So
[tex]f(x)=\displaystyle\int(-x)\,\mathrm dx=-\dfrac{x^2}2+C[/tex]
for some constant [tex]C[/tex]. Given that [tex]f(0)=7[/tex], we find that
[tex]7=-\dfrac{0^2}2+C\implies C=7[/tex]
so that on [-3, 0] we have
[tex]f(x)=-\dfrac{x^2}2+7[/tex]
and
[tex]f(-3)=\dfrac52=2.5[/tex]
