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# Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval. $e^x = 3 - 2x$, $(0, 1)$

## The equation $e^{x}=3-2 x$ is equivalent to the equation $e^{x}+2 x-3=0 . f(x)=e^{x}+2 x-3$ is continuous on the interval$[0,1], f(0)=-2,$ and $f(1)=e-1 \approx 1.72 .$ since $-2 < 0 < e-1,$ there is a number $c$ in (0,1) such that $f(c)=0$ by theIntermediate Value Theorem. Thus, there is a root of the equation $e^{x}+2 x-3=0,$ or $e^{x}=3-2 x,$ in the interval (0,1)

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##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

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### Video Transcript

okay, we have an equation E. To the X equals three minus two X. And we want to use the intermediate value theorem to show uh that there is a root of this equation. In other words, we want to show That there is some value of X between zero and 1. Uh that will actually be a solution to this equation. So there's some number for X between zero and one. Uh That could be plugged in for X and will make this equation trip. Now, what I'd like to do first is I'd like to have an equation that says something equal zero, and we can get that simply by adding the two X to both sides of the equation. So I would have E. T. D X plus two, X equals three. I simply got that by adding two extra both sides of the equal sign. And now, so that I can have it say equal zero. Let's subtract three from both sides. So finally, an equivalent equation reads E T D x Plus two, X -3 equals zero. So this is the equation that we need to show has a solution. There is a number for ex uh that is a solution to this equation, and that number for X is in the interval from 01. And we are going to use the intermediate value theorem uh to help us find that. And the way we do that is uh here is our expression E T x plus two, X minus three. I want to show that let's call this uh function F a bex. Well let's let's not get involved with F. A bex. But I want to show that when ex uh if I use zero and plug it into this function, let's see what happens. Okay, when X zero E 20 Plus two times 0 minus three basically will be ah well E 2 to 0, anything to zero is one, two times 0 is zero. So 1 Plus 0 to track three is -2. So when X0, the expression evaluated for X equals zero comes out to be negative too. Okay so we're trying to find a we're trying to show that there has to be some number X between zero and one that makes this equation equal to zero. We just showed that when X is zero itself, this expression is negative, emphasis on negative too. Now let's see what happens when X is one. When X is one, what does this expression on the left side equal sign. Come out to B E T T X plus two X minus three. What is that when we plug one in for X? Well, ETDX would be easy to the one Plus two times x plus two times 1 minus straight. Well, E to the first is just E plus two times one, which is two minus three. E plus two minus three is E minus one. E -1. E. is approximately the number 2.71. So E -1 is approximately 2.7. -1. or approximately Positive 1.71, emphasis on the positive. So what have we shown so far? We have shown that when X0 this expression is negative. When X is one this expression is positive. So when X0, okay our function our expression is negative, but when X is one hour expression is positive. So if our expression okay let's let's put like a little little box around it. This expression is negative where next zero this expression is positive when X is one. So when X zero this is negative. When X is one, this is positive. So we're going from this expression being negative at X equals zero to this expression being positive at X equals one. So when we move when X moves from 0 to 1, this expression moves from being negative to being positive. Well if this expression moves from being negative to positive at some point it must have been zero. If you're going from negative territory into positive territory you have to pass zero. So there is some value for X. And this is what the intermediate value theorem tells us. Uh Since this expression is negative at zero and this expression is positive at one, there is some value of X between zero and one. Where this expression has to be zero, it starts out negative, it ends up positive somewhere in this interval it had to be zero when it went from negative to positive. So that proves that there must be some value of X in this interval. Uh that is a solution to this equation where this equation equals zero. So in other words, the X number in this interval, that is a solution to this equation. That makes this expression equal zero. That X is a root of this equation. And let's take a little look at demos. I have that expression graft here in dez mose uh keeping X restricted uh to be between zero and one. So that's why you only see the portion of the graph between X equals zero. Xs one Here when X was zero. The function the expression was negative when X was one, it was positive. Since the expression okay, each of the X plus two, X minus three. Since it went from negative to being positive, there must have been some point where it was equal zero. When it goes from negative to positive. There must be some point where it was zero. And you can see on the graph here that we actually found the point where this expression is zero. We actually found the root of the equation. The value of X. That makes this expression equal to zero. When X is .594, the function or the expression is zero

Temple University

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Limits

Derivatives

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

Lectures

Join Bootcamp