Let $f:\mathbb R\to\mathbb R$ be a twice differentiable function. Suppose that for all real numbers $x$ and $y$, the function $f$ satisfies $f'(x)-f'(y) \leq 3|x-y|$
(a) Show that for all $x$ and $y$, we must have $|f(x)-f(y)-f'(y)(x-y)|\leq 1.5(x-y)^{2}$.
(b) Find the largest and smallest possible values for $f''(x)$ under the given conditions.
I think (b) is trivial. The range is $\pm {3}$. I'm looking for example of that function. For (a) I started with substituting $t=x-y$ assuming $x>y$ but i have no idea about about range of $t$ and how to approach further.
Edit- As mentioned in comment one example of (b) is $f(x)=1.5x^{2}$.
(a) is false. Let $f(x) = 1.5x^2 + 1.$ Then $f'$ satisfies the condition. But if $x=y= 1,$ the left side of the alleged inequality is $|f(1)-f(1) -3f(0)|=3,$ while the right side is $0.$
I'm guessing the inequality in (b) should be
$$|f(x)-f(y)-f'(y)(x-y)|\leq 1.5(x-y)^{2}.$$