I don't know how this problem is solved.

N

nombreuso

New member
Joined
Feb 26, 2021
Messages
35
What value for the constant c will make the function e^(-x) + 1 + cx approximately constant, for values of x near 0?
Find the function y(x) satisfying = xy + x, y(0) = 1
These are two problems, can you solve one of them as an example?
 
nombreuso said:
What value for the constant c will make the function e^(-x) + 1 + cx approximately constant, for values of x near 0?
Find the function y(x) satisfying = xy + x, y(0) = 1
These are two problems, can you solve one of them as an example?
Click to expand...
1. What can we say about the derivative of this function at 0?
2. =xy+x?
 
For 1. do you know the expansion of [MATH]e^x[/MATH]?
[MATH]e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...[/MATH]so [MATH]e^{-x}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+...[/MATH]
[MATH]\therefore [/MATH] near [MATH]x=0[/MATH], [MATH]e^{-x} \approx 1-x[/MATH]so what value of c would make [MATH]1-x+1+cx[/MATH] approx constant?

For 2. you need to restate this because it is not clear what it means.

lev888's method for question 1 is nicer than mine.
 
Sorry, I pasted another exercise by accident, so I'll focus on the first exercise only. It's supossed to be a problem about approximation, so i wrote e^(-x) + sqrt(1 + cx) is approximately (1+x)(1+cx/2) = 1 + (c/2 +1)x + (cx^2)/2
 
nombreuso said:
Sorry, I pasted another exercise by accident, so I'll focus on the first exercise only. It's supossed to be a problem about approximation, so i wrote e^(-x) + sqrt(1 + cx) is approximately (1+x)(1+cx/2) = 1 + (c/2 +1)x + (cx^2)/2
Click to expand...
This makes your question more obscure rather than less.

The initial question seemed to be

[MATH]\text {What is } c \text { if } f(x) = e^{-x} + 1 + cx \text { and } f(x) \approx k \text { when } x \approx 0.[/MATH]
The question seems almost trivial because, as lev so strongly hinted,

[MATH]\left (\dfrac{d}{dx} k \right ) = 0 \text { and } f’(x) = - \dfrac{1}{e^x} + c \approx c - 1 = 0 \text { when } x \approx 0.[/MATH]
For what value of c does c - 1 = 0?

Lex reaches the same point by a perhaps more rigorous but less intuitive (to me at least) route.

Based on your second post, I am now not even sure what the actual question is. If the wrong exercise was posted, which is the first exercise in the picture you did not show us in either post.
 
Last edited:
What value for the constant c will make the function e^(-x) + 1 + cx approximately constant, for values of x near 0?
JeffM said:
This makes your question more obscure rather than less.

The initial question seemed to be

[MATH]\text {What is } c \text { if } f(x) = e^{-x} + 1 + cx \text { and } f(x) \approx k \text { when } x \approx 0.[/MATH]
The question seems almost trivial because, as lev so strongly hinted,

[MATH]\left (\dfrac{d}{dx} k \right ) = 0 \text { and } f’(x) = - \dfrac{1}{e^x} + c \approx c - 1 = 0 \text { when } x \approx 0.[/MATH]
For what value of c does c - 1 = 0?

Lex reaches the same point by a perhaps more rigorous but less intuitive (to me at least) route.

Based on your second post, I am now not even sure what the actual question is. If the wrong exercise was posted, which is the first exercise in the picture you did not show us in either post.
Click to expand...
What value for the constant c will make the function e^(-x) + sqrt(1 + cx) approximately constant, for values of x near 0? I made a mistake when pasting the question and don't know how to edit it. Now I'm also confused.
 
nombreuso said:
What value for the constant c will make the function e^(-x) + 1 + cx approximately constant, for values of x near 0?

What value for the constant c will make the function e^(-x) + sqrt(1 + cx) approximately constant, for values of x near 0? I made a mistake when pasting the question and don't know how to edit it. Now I'm also confused.
Click to expand...
Lev’s hint still applies.

If y is a constant, then dy/dx = 0, right?

Locally a differentiable function is approximately a straight line with a slope equal to the derivative. You understand that, right?

So, if the function is almost constant, it looks locally like a horizontal line. And the slope of that line is about zero.

[MATH]f’(x) = - \dfrac{1}{e^x} + \dfrac{c}{2\sqrt{1 + cx}} \approx \dfrac{c}{2} - 1 \text { when } x \approx 0.[/MATH]
Can you solve (c / 2) - 1 = 0?

Make sense?
 
For 1. do you know the expansion of [MATH]e^x[/MATH]?
[MATH]e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...[/MATH]so [MATH]e^{-x}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+...[/MATH]
[MATH]\therefore [/MATH] near [MATH]x=0[/MATH], [MATH]e^{-x} \approx 1-x[/MATH]
[MATH]\sqrt{1+cx}=(1+cx)^\frac{1}{2}\approx 1+\frac{1}{2} cx \hspace2ex \text{ near }x=0[/MATH]so [MATH]e^{-x}+\sqrt{1+cx} \:\approx \: 1-x \hspace2ex + \hspace2ex 1+\frac{1}{2}cx[/MATH]What value of [MATH]c[/MATH] would make [MATH]\hspace2ex 2+\left(\frac{1}{2}c-1\right)x \hspace2ex [/MATH] constant?
 
lex said:
For 1. do you know the expansion of [MATH]e^x[/MATH]?
[MATH]e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...[/MATH]so [MATH]e^{-x}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+...[/MATH]
[MATH]\therefore [/MATH] near [MATH]x=0[/MATH], [MATH]e^{-x} \approx 1-x[/MATH]
[MATH]\sqrt{1+cx}=(1+cx)^\frac{1}{2}\approx 1+\frac{1}{2} cx \hspace2ex \text{ near }x=0[/MATH]so [MATH]e^{-x}+\sqrt{1+cx} \:\approx \: 1-x \hspace2ex + \hspace2ex 1+\frac{1}{2}cx[/MATH]What value of [MATH]c[/MATH] would make [MATH]\hspace2ex 2+\left(\frac{1}{2}c-1\right)x \hspace2ex [/MATH] constant?
Click to expand...
c=2. Thanks, I think I understood.
 
You must log in or register to reply here.
Top