Finding root ,don't get how e^x=2-x work

[Demonfruzz]

(c) By sketching suitable graphs, show that the equation e^x = 2 - x has exactly one root in the interval 0 < x < 2. Use the initial value x₁ = 0.5 to approximate the root of e^x = 2 - x, correct to four decimal places by using Newton's method.

I know how to do the question given y= x

but in this case e^x=2-x what does that mean

do I draw two graph find y intercept in which lies on 2 isint that out of the domain x= (0,2)

Thanks

 

[Ishuda]

(c) By sketching suitable graphs, show that the equation e^x = 2 - x has exactly one root in the interval 0 < x < 2. Use the initial value x₁ = 0.5 to approximate the root of e^x = 2 - x, correct to four decimal places by using Newton's method.

I know how to do the question given y= x

but in this case e^x=2-x what does that mean

do I draw two graph find y intercept in which lies on 2 isint that out of the domain x= (0,2)

Thanks

Possibly the easiest way is to just re-write the equation as
y = e^x + x - 2
and find the x intercept, i.e. y=0

 

[Deleted member 4993]

(c) By sketching suitable graphs, show that the equation e^x = 2 - x has exactly one root in the interval 0 < x < 2. Use the initial value x₁ = 0.5 to approximate the root of e^x = 2 - x, correct to four decimal places by using Newton's method.

I know how to do the question given y= x

but in this case e^x=2-x what does that mean

do I draw two graph find y intercept in which lies on 2 isint that out of the domain x= (0,2)

Thanks

Another graphical way:

plot y = e^x and plot y = 2 - x on the same graph. Now look for points of intersection.

 

[stapel]

(c) By sketching suitable graphs, show that the equation e^x = 2 - x has exactly one root in the interval 0 < x < 2. Use the initial value x₁ = 0.5 to approximate the root of e^x = 2 - x, correct to four decimal places by using Newton's method.

I know how to do the question given y= x

but in this case e^x=2-x what does that mean

It means that you've been given an equation, and they're wanting you to find the x-value(s) for which the equation is true. Back in algebra, you did this by solving equations for the value(s) of x. In this case, they're suggesting that algebraic methods will not work, so you'll need to use other methods.

To start, it might be helpful to think about what you learned back in algebra. When you solved an equation of the form "(something with x in it) = (some number)", you tried to find the x-value(s) that made the (something with x in it) equal to (that number), which was the y-height of the graph. Another way of looking at it was to subtract the (some number) into the (something with x in it), leaving zero on the other side. Then you'd try to find the x-intercepts, which were where y was equal to 0.

In this case, you are given specific instructions, so try following them. Just like in the examples they did in your book and in class, you are given an equation, and asked to solve. Since any solutions are where the two sides of the equation take on the same value, so you're finding any x-value(s) for which e^x equals 2 - x. To "sketch suitable graphs", state these two sides explicitly as functions: y1 = e^x and y2 = 2 - x. (I'm using the naming convention they taught you back in algebra, for plugging these into your graphing calculator.) Then you'll find any intersection points, which is where the lines are "equal" (that is, have the same x- and y-values).

So begin your solution by following the instructions, using the method they outlined in class: treat each side of the equation as its own function; graph the two functions; find where they intersect, at least approximately. This shows that there must be a solution. Then follow the instructions about using Newton's Method. (The graph is helpful in letting you know what's a good first-guess value for the Method.)

If you get stuck, please reply showing your efforts so far. Or, if you remain unable to start, please reply showing a sample worked exercise from your book, stating clearly where it stops making sense to you. Thank you!