Problem

(a) Prove that the equation has at least one real…

03:26

Problem 59 Hard Difficulty

(a) Prove that the equation has at least one real root.
(b) Use your graphing device to find the root correct to three decimal places.

$ 100e^{-x/100} = 0.01x^2 $

Answer

(a) Let $f(x)=100 e^{-x / 100}-0.01 x^{2} .$ Then $f(0)=100>0$ and
$f(100)=100 e^{-1}-100 \approx-63.2<0 .$ So by the Intermediate
Value Theorem, there is a number $c$ in (0,100) such that $f(c)=0$
This implies that $100 e^{-c / 100}=0.01 c^{2}$
(b) Using the intersect feature of the graphing device, we find that the
root of the equation is $x=70.347$, correct to three decimal places.

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Video Transcript

Okay. And this problem uh we have the equation 100 times E. To the negative X over 100 power is equal to 0.1 times X squared. And we need to show that this equation has at least one real root, meaning there is value for X. A real number value for X. That will satisfy this equation. Uh And then we're going to go ahead and uh graph this and actually we're gonna prove that this equation uh does have at least one real number solutions by graphing it. But first, in order to graph it, we want to bring this 10.1 expert over here to the left side. So on the left side, currently we have 100 times e. To the negative x over 100 power. And if I subtract the 0.1 X squared from both sides to 0.1 X squared from both sides of the equation uh we will have equals zero. So 100 times E. To the negative X Over 100 minus 0.1 X squared uh equals zero. So to find a solution, we have to find a value of X that satisfies this equation. What we are going to do is we are going to graft this function that I just put in this box, we're going to graft dysfunction. And uh if this function crosses the X axis, if there is an X intercept on the graph of this function, uh then for the x intercept X would obviously be zero And the Y Coordinate would be zero. So to find a solution. Okay, we just need to look for an X intercept. So we're gonna look for an X intercept on the graph. Well, so next we're going to use decimus and we are going to graph of this function. So here is the function entered into does most and at the moment uh it looks like you don't even see the graph. And so what we need to do is let's just zoom out and see if it comes into uh you know, to be visual. So zoom out, zoom out, zoom out and we can see it's starting to show up. So there is the graph of the function of 100 times e to the negative X over 100 minus point or one X squared. Now you can see that this graph does have an X intercept right at this point. Uh So the graph crosses the X axis at this point. So when X equals 70.347, the function value is zero, meaning dysfunction is equal to zero. When x is 70.347. So that is a root of the original equation. If we look at this original equation, let's go ahead and circle it in green To find a route of this equation meant to find a solution and that solution, the root of that equation is X equals 70.347. So x equals 70 .347. That is a root or solution to this equation, because when X equals this number, uh this expression here was equal to zero.

Leon D.

Temple University

Related Courses

Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 2

Limits and Derivatives

Section 5

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