Answer to Question #223132 in Differential Equations for Ateufack Zeudom

Question

Study the following functions

5) f(x) = I2x - 3Ie^-1/x

Answer 1

1. Domain ( f)=$(-\infty,\space \infty)$ \{0};
2. Vertical asymptote x=0
3. Unilateral limits near x=0

$\lim_{x\to +0} f(x) =|2\cdot 0-3|\cdot e^{\frac{-1}{+0}}=3\cdot e^{-\infty}=3\cdot 0=0$

$\lim_{x\to -0} f(x) =|2\cdot 0-3|\cdot e^{\frac{-1}{-0}}=3\cdot e^{\infty}=3\cdot \infty=\infty$

4 Oblique asymptote

$f(x)=\begin{cases} (2x-3)\cdot e^{-\frac{1}{x}},\space x\ge1.5\\(3-2x)\cdot e^{-\frac{1}{x}},\space x<1.5\ \end{cases}$

$\lim_{|x|\to \pm \infty} e^{\frac{-1}{x}}=e^{-\frac{1}{\pm \infty }}=e^0=1$

so

y=2x-3 - right oblique asymptote

y=3-2x -left oblique asymptote

5 Zeros

f(x)=|2x-3|$\cdot e^{-\frac{1}{x}}=0 \iff x=1.5$

{1.5} - set of zeros of the function;

6 Sign

|2x-3| $\ge 0,\space e^{-\frac{1}{x}}\ge 0\implies f(x) \ge 0$ everywhere

5 Derivative

1) x>1.5 f'(x)=$\left( (2x-3)\cdot e^{\frac{-1}{x}} \right)'=2\cdot e^{-\frac{1}{x}}+(2x-3)\cdot e^{-\frac{1}{x}}\cdot \left( \frac{-1}{x} \right)'=\frac{2x^2+2x-3}{x^2}\cdot e^\frac{-1}{-x}$

2.x<1.5 f'(x)=$\left( (3-2x)\cdot e^{\frac{-1}{x}} \right)'=2\cdot e^{-\frac{1}{x}}+(2x-3)\cdot e^{-\frac{1}{x}}\cdot \left( \frac{-1}{x} \right)'=-\frac{2x^2+2x-3}{x^2}\cdot e^\frac{-1}{-x}$

5.1 Zeros of derivative

2x²+2x-3=0;

$x_{1,2}=\frac{-2 \pm\sqrt{4+24}}{4}=\frac{-1\pm \sqrt 7}{2}$

both values are less than 1.5

5.2 Sign of derivative and monotony, extremes

$x \in \left ( -\infty , \frac{-1-\sqrt 7}{2} \right)$ f'(x)<0 f(x) decreasing

$x \in \left ( \frac{-1-\sqrt 7}{2} , 0 \right ) \cup \left(0,\frac{-1+\sqrt 7}{2} \right)$ f'(x)>0 f(x) increaing

$x \in \left(\frac{-1+\sqrt 7}{2},\space 1.5 \right)$ f'(x)<0 f(x) decreasing

$x\in (1.5, \infty)$ f'(x)>0 f(x) increasing

$x_{1}= \frac{-1-\sqrt 7}{2}$ - point of minimum

$x_{1}= \frac{-1+\sqrt 7}{2}$ - point of local maximum

6 Second derivative

1) x>1.5

f''(x)=$\left((2+\frac{2}{x}-\frac{3}{x^2})\cdot e^{-\frac{1}{x}}\right)'=\left( -\frac{2}{x^2} +\frac{6}{x^3}\right) \cdot e^{-\frac{1}{x}}+(2+\frac{2}{x}-\frac{3}{x^2})\cdot\left( e^{-\frac{1}{x}} \right)'=$

$\left((2+\frac{2}{x}-\frac{3}{x^2})\cdot e^{-\frac{1}{x}}\right)'=\left( -\frac{2}{x^2} +\frac{6}{x^3}\right) \cdot e^{-\frac{1}{x}}+(2+\frac{2}{x}-\frac{3}{x^2})\cdot e^{-\frac{1}{x}}\cdot \frac {1}{x^2}=\\ =\frac{-2x^2+6x+2x^2+2x-3}{x^4}\cdot e^{-\frac{1}{x}}=\frac{8x-3}{x^4}\cdot e^{-\frac{1}{x}}$

2) x<1.5

f''(x)=$-\frac{8x-3}{x^4}\cdot e^{-\frac{1}{x}}$

6.1 Zeros of second derivative

f''(x)=0 $\iff x=\frac{3}{8}<1.5$

Sign of f''(x)

$x \in (-\infty, 0)$ f''(x)>0, f(x) is convex

$x\in(0,\frac{3}{8})$ f"(x)>0 f(x) is convex

$x\in(\frac{3}{8},1.5)$ f''(x)<0 f(x) is concave

$x\in (1.5,\infty)$ f''(x)<0 f(x) ic convex

x_3,4=3/8,1.5 - points d'inflexion

Scetch of graph