Calculus: Differentiable Function and Tangent Line

Question

I am not sure how to approach this question:

Let $f(x)$ be a continuous and differentiable function of order 2. Let $f ''(x)>0$ for all values of $x$. The tangent line to the function at $x=1$ is $y=-x+1$. Show that $f(1.1)>-0.1$.

Thanks!

Answer 1

By doing a second derivative test, We could conclude that $f(x)$ is concave upward since $f''(x)>0$. The only intersection between the tangent line $y=-x+1$ and $f(x)$ should be at $x=1$. And since $f(x)$ is concave upward, the tangent line must lie below $f(x)$ everywhere except at $x=1$.

The y value of the tangent line at $x=1.1$ is $y=-1.1+1=-0.1$. Since $f(x)$ is above that line, $f(1.1)>-0.1$ for sure.

Answer 2

By approximatin f(x) by using two of terms of its Taylor series you can determine the equation of the tangent line as: \begin{equation} y=f(x_0)+f'(x_0)(x-x_0) \end{equation} where $x_0=1$. Then since \begin{equation} f(1)+f'(1)(x-1)= -x+1 \end{equation} we can conclude that $f(1)=0$ and $f'(1)=-1$. When consider the term of order two in the taylor expansion, since $f''(x)$ is always positive, we can conclude that \begin{equation} f(1.1)=f(1)+f'(1)(1.1-1)+ a positive term \end{equation} Then $f(1.1)=-0.1$ + a positive term