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Trigger12 September 20th, 2018 10:32 AM

Stuck at solving equation
 
Hi.

I can't find an answer to the equation $\displaystyle e^x+x-3=0$. I'm not sure how to solve it.

SDK September 20th, 2018 10:44 AM

Just apply Newton's method. This gives a solution $x = .7921...$. Since the expression is monotone increasing, this must be a unique solution.

Trigger12 September 20th, 2018 10:50 AM

I have already done that in the second question where they tell you to use Newton´s method. But the first question is: Show that we can find the value of a, solving the equation e^x+x-3=0.

Trigger12 September 20th, 2018 10:53 AM

The task is:
The tangent to the curve $\displaystyle y=e^{2-x} $ in $\displaystyle x=2 $ cuts the curve $\displaystyle y=e^x $ in a point where x=a.

mathman September 20th, 2018 01:14 PM

The equation has no analytic solution, only numerical.

Trigger12 September 20th, 2018 01:37 PM

Ok, but how to I proceed to solve it numerical?

JeffM1 September 20th, 2018 01:37 PM

Quote:

Originally Posted by Trigger12 (Post 599312)
I have already done that in the second question where they tell you to use Newton´s method. But the first question is: Show that we can find the value of a, solving the equation e^x+x-3=0.

Have you thought about applying the mean value theorem to the function

$f(x) = e^x + x - 3.$

Is f(x) continuous?

What is the sign of f(0)?

What is the sign of f(1)?

What does all that imply about the existence of a such that f(a) = 0?

Trigger12 September 20th, 2018 01:38 PM

Quote:

Originally Posted by JeffM1 (Post 599323)
Have you thought about applying the mean value theorem to the function

$f(x) = e^x + x - 3.$

Is f(x) continuous?

What is the sign of f(0)?

What is the sign of f(1)?

What does all that imply about the existence of a such that f(a) = 0?

Ahh! I didn't think about solving it that way.

JeffM1 September 20th, 2018 01:46 PM

Quote:

Originally Posted by Trigger12 (Post 599324)
Ahh! I didn't think about solving it that way.

Without understanding the exact language of the problem, I cannot be sure my suggestion is even partially relevant. And my Swedish is non-existent so you are on your own in terms of judging how helpful the suggestion is.

Trigger12 September 20th, 2018 01:55 PM

Quote:

Originally Posted by JeffM1 (Post 599325)
Without understanding the exact language of the problem, I cannot be sure my suggestion is even partially relevant. And my Swedish is non-existent so you are on your own in terms of judging how helpful the suggestion is.

In the second question, I used Newton's method to find an approximate value of a, and got 0,792059, that is correct. My problem is that they want us to find the same value, but by solving the equation given over in the first question.


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