1.The minimum of f(x)=∫0xex−tf′(t)dt−(x2−x+1)exf(x)=\int_0^x e^{x-t}f'(t)dt-(x^2-x+1)e^xf(x)=∫0xex−tf′(t)dt−(x2−x+1)ex, x∈Rx\in\mathbb{R}x∈R, isa.−2/e-2/\sqrt{e}−2/eb.−2e-2\sqrt{e}−2ec.−e-\sqrt{e}−ed.2/e2/\sqrt{e}2/eLogin to continueOnly logged in users canattempt or see the solution.