Let f(x)=(x+1)/(xe^(x)+e^(x)) when x!=-1. f is continuous for all real numbers. Find f(-1). Choose 1 answer: (A) -2e (B) 1 (c) e (D) 2

Question

Let  f(x)=(x+1)/(xe^(x)+e^(x)) when  x!=-1.  f is continuous for all real numbers. Find  f(-1). Choose 1 answer: (A)  -2e (B) 1 (c)  e (D) 2

Bytelearn · Accepted Answer

Problem Statement: We are asked to find the value of the function f(x) at x = -1. However, the function is not defined at x = -1, so we need to check if the function has a limit as x approaches -1. If the limit exists, then by the continuity of the function, the limit will be the value of f(-1).Checking for Limit Existence: Let's find the limit of f(x) as x approaches -1. We will use the fact that e^x is continuous everywhere and that the limit of a quotient is the quotient of the limits, provided the limit of the denominator is not zero. lim (x -> -1) f(x) = lim (x -> -1) (x+1)/(xe^(x)+e^(x))Using the Continuity of e^x: We substitute x = -1 into the numerator and denominator to see if we can directly evaluate the limit. lim (x -> -1) (x+1)/(xe^(x)+e^(x)) = (-1+1)/((-1)e^(-1)+e^(-1))Substituting x = -1: Simplify the expression by performing the arithmetic operations. 0/((-1)e^(-1)+e^(-1)) = 0/(-e^(-1)+e^(-1))Simplifying the Expression: We notice that the terms in the denominator cancel each other out because they are equal and opposite. 0/(0) is undefined, so we cannot directly evaluate the limit this way. We need to find another approach to determine the limit as x approaches -1.Applying L'Hôpital's Rule: We can apply L'Hôpital's Rule because we have an indeterminate form of 0/0. L'Hôpital's Rule states that if the limit as x approaches a of f(x)/g(x) is 0/0 or ±∞/±∞, then the limit is the same as the limit of the derivatives of the numerator and denominator, provided that the latter limit exists. lim (x -> -1) f(x) = lim (x -> -1) (d/dx(x+1))/(d/dx(xe^(x)+e^(x)))Finding the Derivatives: We find the derivatives of the numerator and the denominator. The derivative of the numerator (x+1) with respect to x is 1. The derivative of the denominator (xe^(x)+e^(x)) with respect to x is e^(x) + xe^(x) + e^(x) using the product rule and the fact that the derivative of e^(x) is e^(x).Substituting the Derivatives: Now we substitute the derivatives back into the limit. lim (x -> -1) f(x) = lim (x -> -1) 1/(e^(x) + xe^(x) + e^(x))Simplifying the Denominator: We simplify the denominator by combining like terms. lim (x -> -1) 1/(e^(x) + xe^(x) + e^(x)) = lim (x -> -1) 1/(e^(x)(1 + x + 1))Evaluating the Limit: Now we substitute x = -1 into the simplified expression to evaluate the limit. lim (x -> -1) 1/(e^(x)(1 + x + 1)) = 1/(e^(-1)(1 - 1 + 1))Final Result: Simplify the expression by performing the arithmetic operations in the denominator. 1/(e^(-1)(1)) = 1/(e^(-1)) = eWe have found the limit of f(x) as x approaches -1, which is e. Since the function f is continuous for all real numbers except x = -1, and we have found the limit as x approaches -1, we can conclude that f(-1) = e.

Let $f(x)=\frac{x+1}{x e^{x}+e^{x}}$ when $x \neq-1$ . $\newline$ $f$ is continuous for all real numbers. $\newline$ Find $f(-1)$ . $\newline$ Choose $1$ answer: $\newline$ (A) $-2 e$ $\newline$ (B) $1$ $\newline$ (C) $e$ $\newline$ (D) $2$

Full solution

More problems from Domain and range of quadratic functions: equations

Related topics

Let f(x)=x+1xex+ex f(x)=\frac{x+1}{x e^{x}+e^{x}} f(x)=xex+exx+1​ when x≠−1 x \neq-1 x=−1.\newlinef f f is continuous for all real numbers.\newlineFind f(−1) f(-1) f(−1).\newlineChoose 111 answer:\newline(A) −2e -2 e −2e\newline(B) 111\newline(C) e e e\newline(D) 222

Let $f(x)=\frac{x+1}{x e^{x}+e^{x}}$ when $x \neq-1$ . $\newline$ $f$ is continuous for all real numbers. $\newline$ Find $f(-1)$ . $\newline$ Choose $1$ answer: $\newline$ (A) $-2 e$ $\newline$ (B) $1$ $\newline$ (C) $e$ $\newline$ (D) $2$