The problem, encountered in a high school math textbook in the exercises on the MVT,goes as following:
Let $f(x)=e^x-ex, x\geq 0$ and $f(\ln{2})<2$, prove that $y=(1-e)x+1$ is tangent to the graph of $f, C_f,$ and find $b, c \in [0,+\infty)$ such that $$f(b)+f(c-2)\leq (1-e)(b+c-2)+2$$ holds.
I have started by finding where $y$ is tangent to $f(x)$, so it must be $y'=f'(x)\Rightarrow 1-e=e^x-e\Rightarrow x=0$. So it must be tangent at point $A(0, 1)$. My first thought was breaking down the inequality to obtain $f(b)+f(c-2)\leq y(b)+y(c-2)$. Could I perhaps use the Mean Value Theorem on a new function $g(x)=f(x)-y(x)$ in the interval $(0,1)$? Any help would be appreaciated!
What a weird, contrived and unnecessary problem. Anyway, if this statement does not have an error, then you it must be the case that $b = c - 2 = 0.$ That's because, your function $f(x)$ is convex on the entire real line, meaning that it's graph lies above any tangent line i.e.,: $$f(x)\geq y(x)$$ for all $x$ and the inequality is strict if $x\neq 0.$ Therefore, the only way you can have $$f(b)+f(c-2)\leq y(b)+y(c-2)$$ is if $b = c - 2 = 0. $