How do we know that $\int_1^x \frac{|2+t|}{t^2+1}dt$ has a horizontal tangent in $x=-2$?

Question

Graphing the function $F(x)=\int_1^x \frac{|2+t|}{t^2+1}dt$ I know that it has a horizontal tangent in the point $x=-2$, and that it is an increasing function in $\mathbb{R}$. I was wondering if there was another way in telling the same things without graphing the function.

Answer 1

$F(x)=\int_1^x \frac{|2+t|}{t^2+1}dt$ is differentiable, and $$F'(x)=\frac{|2+x|}{x^2+1}.$$ This implies that $F'(-2)=0$.

Answer 2

For all $x\in\mathbb{R}$ :

$F’(x)=\frac{|2+x|}{1+x^2}$

Hence $F’(-2)=0$, which means that the slope of the tangent line at $(-2,F(-2))$ is zero.

In other words, that tangent is horizontal.

Moreover, $F’(x)\geqslant0$ for all $x$, which proves that $F$ is non decreasing.

Answer 3

You can define $F(x)$ in the following two regions:

$$x > -2$$ $$x \le -2$$

When $x >-2$,

$$F(x) = \int_{1}^{x} \frac{2+t}{1+t^2} dt$$

When $ x \le -2$

$$F(x) = \int_{1}^{-2} \frac{2+t}{1+t^2} dt - \int_{-2}^{x} \frac{2+t}{1+t^2} dt$$

Now calculate $F'(x)$ using Lebnitz Rule and verify that $F'(-2^-) = F'(-2^+) = 0$.

Answer 4

The fundamental theorem of calculus tells us that $F'(x)=\dfrac {\mid 2+x\mid}{x^2+1}$. Then $F'(-2)=0$.