Fonction Exponentielle : Exercices

Correction :

Voici les simplifications :

1. $e^{2x} \cdot e^{3x} = e^{2x + 3x} = e^{5x}$

2. $\frac{e^{5x}}{e^{2x}} = e^{5x - 2x} = e^{3x}$

3. $(e^{-x})^4 = e^{-4x}$

4. $\frac{e^{x+1}}{e^{x-1}} = e^{(x+1) - (x-1)} = e^{x+1-x+1} = e^2$

5. $e^{2x} \cdot e^{-x} \cdot e^3 = e^{2x - x + 3} = e^{x+3}$

Correction :

1. $e^{2x} = e^3 \implies 2x = 3 \implies x = \frac{3}{2}$
2. $e^{x^2} = e^{4x-3} \implies x^2 = 4x-3 \implies x^2 - 4x + 3 = 0$. Factorisons : $(x-1)(x-3) = 0$. Solutions : $x=1$ ou $x=3$.
3. $e^{3x+1} \cdot e^{-x+2} = e^5 \implies e^{3x+1-x+2} = e^5 \implies e^{2x+3} = e^5 \implies 2x+3 = 5 \implies 2x = 2 \implies x = 1$

Correction :

1. $e^{x+2} \leq e^4 \implies x+2 \leq 4 \implies x \leq 2$
2. $e^{-2x+1} > e^3 \implies -2x+1 > 3 \implies -2x > 2 \implies x < -1$
3. $e^{x^2} < e \implies x^2 < 1 \implies -1 < x < 1$

Correction :

1. $\lim_{x \to +\infty} e^{2x} = +\infty$

2. $\lim_{x \to -\infty} e^{-x} = +\infty$

3. $\lim_{x \to 0} \frac{e^{3x}-1}{x} = \lim_{x \to 0} 3\frac{e^{3x}-1}{3x} = 3 \lim_{X \to 0} \frac{e^X-1}{X} = 3$ (en posant $X=3x$)

Correction :

1. $f'(x) = 3e^{3x+1}$

2. $g'(x) = 2xe^{-x} - x^2e^{-x} = (2x - x^2)e^{-x}$

3. $h'(x) = \frac{e^x \cdot x - e^x \cdot 1}{x^2} = \frac{e^x(x-1)}{x^2}$

Correction :

1. $f'(x) = -2xe^{-x^2}$. Le signe de $f'(x)$ est le signe de $-2x$. Donc $f'(x) > 0$ si $x < 0$ et $f'(x) < 0$ si $x > 0$. $f$ est croissante sur $]-\infty; 0]$ et décroissante sur $[0; +\infty[$.

2. $g'(x) = e^x + (x+1)e^x = (x+2)e^x$. Comme $e^x > 0$, le signe de $g'(x)$ est celui de $x+2$. Donc $g$ est décroissante sur $]-\infty; -2]$ et croissante sur $[-2; +\infty[$.

Correction :

1. $\frac{e^{3x} \cdot e^{-2x}}{e^x} = \frac{e^{3x - 2x}}{e^x} = \frac{e^x}{e^x} = e^{x-x} = e^0 = 1$

2. $(e^{2x})^3 \cdot e^{-5x} = e^{6x} \cdot e^{-5x} = e^{6x - 5x} = e^x$

3. $\frac{1}{e^{-x}} + e^x = e^x + e^x = 2e^x$

Correction :

1. $e^{2x-1} = 1 \Leftrightarrow e^{2x-1} = e^0 \Leftrightarrow 2x - 1 = 0 \Leftrightarrow x = \frac{1}{2}$

2. $e^{x^2} = e^{2x} \Leftrightarrow x^2 = 2x \Leftrightarrow x^2 - 2x = 0 \Leftrightarrow x(x-2) = 0$. Solutions : $x=0$ ou $x=2$.

3. $(e^x - 1)(e^{2x} - 4) = 0 \Leftrightarrow e^x - 1 = 0$ ou $e^{2x} - 4 = 0$. - $e^x - 1 = 0 \implies e^x = 1 \implies x = 0$ - $e^{2x} - 4 = 0 \implies e^{2x} = 4 \implies 2x = \ln 4 \implies x = \frac{1}{2}\ln 4 = \ln 2$ Solutions : $x=0$ ou $x=\ln 2$.

Correction :

1. $e^{3x} > e^{-2} \implies 3x > -2 \implies x > -\frac{2}{3}$

2. $e^{x^2} \leq e^{2x} \implies x^2 \leq 2x \implies x^2 - 2x \leq 0 \implies x(x-2) \leq 0$. Solution : $x \in [0; 2]$

3. $(e^x - 1)(e^{2x} - 4) < 0$. On étudie le signe de chaque facteur : - $e^x - 1 > 0 \implies e^x > 1 \implies x > 0$ - $e^{2x} - 4 > 0 \implies e^{2x} > 4 \implies 2x > \ln 4 \implies x > \ln 2$ On fait un tableau de signes :