Point of intersection between two graphs

Question

I need the points of intersection between these two graphs: f(x)=5e^-x and g(x)=(-e^x)+6. I am kind of stuck with them. 

 

Thank you!

Answer 1

To find the points of intersection, set f(x) = g(x).

 

5e^-x = 6 - e^x

 

5/e^x  = 6 - e^x

 

Multiply by e^x to obtain 5 = 6e^x - (e^x)²

 

Let u = e^x to get u² - 6u + 5 = 0

 

                        (u-5)(u-1) = 0

 

                                     u = 5 or u = 1

 

So, e^x = 5 or e^x = 1

 

x = ln5 or 0

 

Points of intersection:  (ln5, f(ln5)) and (0, f(0))

 

                            = (ln5, 1) and (0, 5)

 

Answer 2

The points of intersection between the two graphs should be all the coordinates containing all values of x for which the 2 graphs are equal.

 

So, the points of intersection are the coordinates (x,y) for all x such that

5e^(-x) = (-e^x) + 6

 

So, let's solve for x.

 

5e^(-x) is the same as 5 * 1/(e^x).

So, 5e^(-x) = 5/(e^x)

 

5/(e^x) = -e^x + 6; let's get all our x's on one side...

 

5/(e^x) + e^x = 6; let's get a common denominator, so that we can get all our e^x's as close as we can.

 

5/(e^x) ^ (e^2x)/(e^x) = 6;

[5+(e^2x)]/[e^x] = 6;

 

Hey, maybe we should get that e^x out of the denominator... Let's multiply both sides by e^x.

 

5 + e^2x = 6e^x; If we move 6e^x to the left side, we have a quadratic type equation.

 

e^2x - 6e^x + 5 = 0; if we substitute 3^x for a, we have...

a^2 - 6a + 5 = 0; this factors nicely into...

(a - 5)(a - 1) = 0; so when a = 5 or a = 1, the graphs intersect. But remember, a is e^x; we still have not found our x values.

 

If, e^x = 5; ln(e^x) = ln5, and the natural log of e^x is x. So, x = ln 5.

If, e^x = 1; x must equal ln 1.

 

So, our x coordinates are ln 1 & ln 5; if we plug them back into either equation, we will get the corresponding y coordinates.

 

I'll use f(x);

 

1. If x = ln1, 5e^-x = 5e^(-ln1);

Remember that -ln1 = ln (1^-1) = ln (1/1) = ln 1; also, remember that e^lnx = x. So, 5e^(-ln1) = 5e^(ln1) = 5*1 = 5. Therefore, one of the points of intersection is the point (ln1, 5).

 

2. If x = ln 5, 5e^-x = 5e^(-ln5);

-ln5 = ln^(5^-1) = ln(1/5); 5e^(ln[1/5]) = 5*(1/5) = 1; So, the other point of intersection is the point (ln 5, 1).

 

So, to sum up (you can skip this if you want)...

The points of intersection for two graphs are the points at which their equations are equal.

 

We,

1. simplified the negative exponent on the left side,

2. moved the (-e^x) to the left side,

3. combined the two terms into one fraction,

4. got the (e^x) out of the denominator,

4. noticed the equation looked quadratic-esque,

5. proceeded to factor,

6. solved for x, remembering our properties of logarithms.

 

Then, after this point we plugged in the x values into one of the equations, solving for the corresponding y values.

 

Note: since we found the x values for which the 2 equations are equal, it doesn't matter which equation you plug the x values back into. You will get the same y values.