Let $ f(x) = x^3+3x+2 $ and $ x=c $ be a point such that $(b-a)f'(c)$ is not equal to $f(b)-f(a)$ for any two values of $a,b\in \mathbf R$ then the number of such points are?
My answer: I thought there will be no such points according to LMVT but the solution says that for such points $x=c , f''(c)=0$ And therefore there will be one such $c.$ But I don't get it.
On
Look at the graph. If you draw a tangent at the inflection point, it will lie below the graph on the right and above the graph on the left. No line parallel to it intersects the graph twice.
I'm just reasoning from the picture. I haven't proved it. Why don't you try?
For a fixed $c$, consider the function
$$h_c(a,b):=f(b)-f(a)-(b-a)f'(c)=f(b)-f(a)-(b-a)(3c^2+3).$$
Notice that if you fix $a$, then $\lim_{b\rightarrow\infty}h_c(a,b)=\infty$ and $\lim_{b\rightarrow-\infty}h_c(a,b)=-\infty$. Since $h$ is continuous in $b$, this means that by IVT there must be a point $b=b_c(a)$ such that $h_c(a,b_c)=0$. So there is no such $c$.