CARACTÉRISATION DE LA FONCTION EXPONENTIELLE
CHARACTERIZATION OF THE EXPONENTIAL FUNCTION
CARACTERIZACIÓN DE LA FUNCIÓN EXPONENCIAL
Exercices
Exercises
Ejercicios
Exercice 1 : Simplification d'expressions
Exercise 1: Simplifying expressions
Ejercicio 1: Simplificación de expresiones
En utilisant les propriétés de la fonction exponentielle, calculer ou simplifier chacune des expressions suivantes :
$e^4 \times e^{-2}$
$\dfrac{e^6}{e^2}$
$e^5 \times e^{-5}$
$\left(e^2\right)^3$
$\dfrac{e^4 \times e^3}{e^5}$
$e^1 + e^0$
Using the properties of the exponential function, calculate or simplify each of the following expressions:
$e^4 \times e^{-2}$
$\dfrac{e^6}{e^2}$
$e^5 \times e^{-5}$
$\left(e^2\right)^3$
$\dfrac{e^4 \times e^3}{e^5}$
$e^1 + e^0$
Usando las propiedades de la función exponencial, calcula o simplifica cada una de las siguientes expresiones:
$e^4 \times e^{-2}$
$\dfrac{e^6}{e^2}$
$e^5 \times e^{-5}$
$\left(e^2\right)^3$
$\dfrac{e^4 \times e^3}{e^5}$
$e^1 + e^0$
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Corrigé :
Solution:
Solución:
$e^4 \times e^{-2} = e^{4 - 2} =$ $e^2$
$\dfrac{e^6}{e^2} = e^{6 - 2} =$ $e^4$
$e^5 \times e^{-5} = e^{5 - 5} = e^0 =$ $1$
$\left(e^2\right)^3 = e^{2 \times 3} =$ $e^6$
$\dfrac{e^4 \times e^3}{e^5} = \dfrac{e^{4+3}}{e^5} = \dfrac{e^7}{e^5} = e^{7-5} =$ $e^2$
$e^1 + e^0 =$ $e + 1$
$e^4 \times e^{-2} = e^{4 - 2} =$ $e^2$
$\dfrac{e^6}{e^2} = e^{6 - 2} =$ $e^4$
$e^5 \times e^{-5} = e^{5 - 5} = e^0 =$ $1$
$\left(e^2\right)^3 = e^{2 \times 3} =$ $e^6$
$\dfrac{e^4 \times e^3}{e^5} = \dfrac{e^{4+3}}{e^5} = \dfrac{e^7}{e^5} = e^{7-5} =$ $e^2$
$e^1 + e^0 =$ $e + 1$
$e^4 \times e^{-2} = e^{4 - 2} =$ $e^2$
$\dfrac{e^6}{e^2} = e^{6 - 2} =$ $e^4$
$e^5 \times e^{-5} = e^{5 - 5} = e^0 =$ $1$
$\left(e^2\right)^3 = e^{2 \times 3} =$ $e^6$
$\dfrac{e^4 \times e^3}{e^5} = \dfrac{e^{4+3}}{e^5} = \dfrac{e^7}{e^5} = e^{7-5} =$ $e^2$
$e^1 + e^0 =$ $e + 1$
Exercice 2 : Calcul de dérivées
Exercise 2: Calculating derivatives
Ejercicio 2: Cálculo de derivadas
Calculer la dérivée $f'(x)$ de chacune des fonctions suivantes, définies et dérivables sur $\mathbb{R}$.
$f(x) = e^x - 5$
$f(x) = 4e^x$
$f(x) = -3e^x + 2x$
$f(x) = e^x - x^2 + 5x - 2$
$f(x) = 2e^x - \dfrac{1}{3}x^3$
Calculate the derivative $f'(x)$ of each of the following functions, defined and differentiable on $\mathbb{R}$.
$f(x) = e^x - 5$
$f(x) = 4e^x$
$f(x) = -3e^x + 2x$
$f(x) = e^x - x^2 + 5x - 2$
$f(x) = 2e^x - \dfrac{1}{3}x^3$
Calcula la derivada $f'(x)$ de cada una de las siguientes funciones, definidas y derivables en $\mathbb{R}$.
$f(x) = e^x - 5$
$f(x) = 4e^x$
$f(x) = -3e^x + 2x$
$f(x) = e^x - x^2 + 5x - 2$
$f(x) = 2e^x - \dfrac{1}{3}x^3$
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Corrigé :
Solution:
Solución:
$f'(x) = e^x$ $f'(x) = 4e^x$ $f'(x) = -3e^x + 2$ $f'(x) = e^x - 2x + 5$ $f'(x) = 2e^x - x^2$
$f'(x) = e^x$ $f'(x) = 4e^x$ $f'(x) = -3e^x + 2$ $f'(x) = e^x - 2x + 5$ $f'(x) = 2e^x - x^2$
$f'(x) = e^x$ $f'(x) = 4e^x$ $f'(x) = -3e^x + 2$ $f'(x) = e^x - 2x + 5$ $f'(x) = 2e^x - x^2$
Exercice 3 : Équations et inéquations
Exercise 3: Equations and inequalities
Ejercicio 3: Ecuaciones e inecuaciones
Résoudre les équations et inéquations suivantes dans $\mathbb{R}$ (on rappelle que $\exp$ est strictement positive sur $\mathbb{R}$) :
$e^x = e$
$e^x \times e^2 = e^5$
$\dfrac{e^{3x}}{e^x} = e^6$
$2e^x = 0$
$e^x + 1 > 0$
$\left(e^x\right)^3 = e^{12}$
Solve the following equations and inequalities in $\mathbb{R}$ (recall that $\exp$ is strictly positive on $\mathbb{R}$):
$e^x = e$
$e^x \times e^2 = e^5$
$\dfrac{e^{3x}}{e^x} = e^6$
$2e^x = 0$
$e^x + 1 > 0$
$\left(e^x\right)^3 = e^{12}$
Resuelve las siguientes ecuaciones e inecuaciones en $\mathbb{R}$ (recuerda que $\exp$ es estrictamente positiva en $\mathbb{R}$):
$e^x = e$
$e^x \times e^2 = e^5$
$\dfrac{e^{3x}}{e^x} = e^6$
$2e^x = 0$
$e^x + 1 > 0$
$\left(e^x\right)^3 = e^{12}$
Voir le corrigé
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Corrigé :
Solution:
Solución:
$e^x = e^1 \iff$ $x = 1$ . L'ensemble des solutions est $S = \{1\}$.
$e^x \times e^2 = e^5 \iff e^{x+2} = e^5 \iff x + 2 = 5 \iff$ $x = 3$ . $S = \{3\}$.
$\dfrac{e^{3x}}{e^x} = e^6 \iff e^{3x-x} = e^6 \iff e^{2x} = e^6 \iff 2x = 6 \iff$ $x = 3$ . $S = \{3\}$.
$2e^x = 0 \iff e^x = 0$. Or pour tout réel $x$, $e^x > 0$. L'équation n'a pas de solution . $S = \emptyset$.
$e^x + 1 > 0 \iff e^x > -1$. Or pour tout réel $x$, $e^x > 0 > -1$. L'inéquation est toujours vraie, donc $S = \mathbb{R}$ .
$\left(e^x\right)^3 = e^{12} \iff e^{3x} = e^{12} \iff 3x = 12 \iff$ $x = 4$ . $S = \{4\}$.
$e^x = e^1 \iff$ $x = 1$ . The solution set is $S = \{1\}$.
$e^x \times e^2 = e^5 \iff e^{x+2} = e^5 \iff x + 2 = 5 \iff$ $x = 3$ . $S = \{3\}$.
$\dfrac{e^{3x}}{e^x} = e^6 \iff e^{3x-x} = e^6 \iff e^{2x} = e^6 \iff 2x = 6 \iff$ $x = 3$ . $S = \{3\}$.
$2e^x = 0 \iff e^x = 0$. But for any real $x$, $e^x > 0$. The equation has no solution . $S = \emptyset$.
$e^x + 1 > 0 \iff e^x > -1$. But for any real $x$, $e^x > 0 > -1$. The inequality is always true, so $S = \mathbb{R}$ .
$\left(e^x\right)^3 = e^{12} \iff e^{3x} = e^{12} \iff 3x = 12 \iff$ $x = 4$ . $S = \{4\}$.
$e^x = e^1 \iff$ $x = 1$ . El conjunto de soluciones es $S = \{1\}$.
$e^x \times e^2 = e^5 \iff e^{x+2} = e^5 \iff x + 2 = 5 \iff$ $x = 3$ . $S = \{3\}$.
$\dfrac{e^{3x}}{e^x} = e^6 \iff e^{3x-x} = e^6 \iff e^{2x} = e^6 \iff 2x = 6 \iff$ $x = 3$ . $S = \{3\}$.
$2e^x = 0 \iff e^x = 0$. Pero para todo número real $x$, $e^x > 0$. La ecuación no tiene solución . $S = \emptyset$.
$e^x + 1 > 0 \iff e^x > -1$. Pero para todo real $x$, $e^x > 0 > -1$. La inecuación siempre es verdadera, por lo que $S = \mathbb{R}$ .
$\left(e^x\right)^3 = e^{12} \iff e^{3x} = e^{12} \iff 3x = 12 \iff$ $x = 4$ . $S = \{4\}$.
Exercice 4 : Suite et exponentielle
Exercise 4: Sequence and exponential
Ejercicio 4: Sucesión y exponencial
Soit $(u_n)$ la suite définie pour tout $n \in \mathbb{N}$ par $u_n = e^{2n+1}$.
Montrer que $(u_n)$ est une suite géométrique dont on précisera le premier terme $u_0$ et la raison $q$.
Calculer les valeurs exactes de $u_0$, $u_1$ et $u_2$.
Let $(u_n)$ be the sequence defined for any $n \in \mathbb{N}$ by $u_n = e^{2n+1}$.
Show that $(u_n)$ is a geometric sequence specifying the first term $u_0$ and the common ratio $q$.
Calculate the exact values of $u_0$, $u_1$ and $u_2$.
Sea $(u_n)$ la sucesión definida para todo $n \in \mathbb{N}$ por $u_n = e^{2n+1}$.
Demuestra que $(u_n)$ es una sucesión geométrica precisando el primer término $u_0$ y la razón $q$.
Calcula los valores exactos de $u_0$, $u_1$ y $u_2$.
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Corrigé :
Solution:
Solución:
Pour tout entier $n \in \mathbb{N}$ : $\dfrac{u_{n+1}}{u_n} = \dfrac{e^{2(n+1)+1}}{e^{2n+1}} = \dfrac{e^{2n+3}}{e^{2n+1}} = e^{(2n+3)-(2n+1)} = e^2$.$q = e^2$ et de premier terme $u_0 = e^{2(0)+1} = e^1 = e$ .
$u_0 = e$ $u_1 = e^{2(1)+1} = e^3$ $u_2 = e^{2(2)+1} = e^5$
For any integer $n \in \mathbb{N}$: $\dfrac{u_{n+1}}{u_n} = \dfrac{e^{2(n+1)+1}}{e^{2n+1}} = \dfrac{e^{2n+3}}{e^{2n+1}} = e^{(2n+3)-(2n+1)} = e^2$.$q = e^2$ and first term $u_0 = e^{2(0)+1} = e^1 = e$ .
$u_0 = e$ $u_1 = e^{2(1)+1} = e^3$ $u_2 = e^{2(2)+1} = e^5$
Para cualquier número entero $n \in \mathbb{N}$: $\dfrac{u_{n+1}}{u_n} = \dfrac{e^{2(n+1)+1}}{e^{2n+1}} = \dfrac{e^{2n+3}}{e^{2n+1}} = e^{(2n+3)-(2n+1)} = e^2$.$q = e^2$ y primer término $u_0 = e^{2(0)+1} = e^1 = e$ .
$u_0 = e$ $u_1 = e^{2(1)+1} = e^3$ $u_2 = e^{2(2)+1} = e^5$