The curve with equation y=4e^x + 6e^-x -11 intersects Xaxis at points A and b, find a and b?

Question

I'm really stuck on this question and have a maths exam tomorrow, and i have been warned a similar question may come up, can anyone explain how to do this?
The curve with equation y=4e^x + 6e^-x -11 intersects Xaxis at points A and b, find a and b

^ repesents 'to the power'

Answer 1

let t=e^x

y=0 when x=a and x=b

substitute t into
y=4e^x + 6e^-x -11

4t+6/t-11=0
multiply by t
4t^2-11t+6=0
using formula
t=2 or 3/4
therefore,at y=0,
e^a=2 or 3/4
{take ln of each side}
>>>>a=ln2 or a=ln(3/4)
at y=0,
e^b=2 or 3/4
{take ln of each side}
>>>>b=ln2 or b=ln(3/4)

therefore,a=ln2=0.6931471806
or
b=ln(3/4)= -0.2876820725
a= ln(3/4)= -0.2876820725
or b= ln2 = 0.6931471806

{check answer by substituting
into original equation
y=4e^(ln3/4)+6/e^(ln3/4)-11
for x=a=ln(3/4)
=4*3/4+6*4/3-11
=3+8-11=0
for x=b=ln2
y=4e^(ln2)+6/e^(ln2)-11
=4*2+6/2-11=0
these two values are correct
at y=0}
we don't know which is a and
which is b,so the two values
are interchangeable

i hope that this helps

Answer 2

Well, lets begin:

The point where the x intercept the axis is when y=0
y= 4e^x+6/e^x -11 = 0

4e^x.e^x+6 - 11e^x = 0
4(e^x)^2 - 11e^x + 6 = 0

If you do: z=e^x you have:

4z^2 - 11z + 6 = 0

This is a second degree equation. You can solve using Baskara:

z = 2 or z = 3/4

Then, if z = e^x

2 = e^x ---> x = ln 2

3/4 = e^x --> x = ln(3/4)

This is the two points where the curve intercept X axis.

For y axis you need x=0. Then:

y = 4e^0+6e^-o - 11
y = 4+6-11

y = -1

This is the point where the curve interceps the Y axis.

Answer 3

When a line intersects the x-axis, the y value must be 0.
y = 4e^x + 6e^-x - 11
0 = 4e^x + 6/e^x - 11
0 = 4(e^x)^2 - 11e^x + 6
e^x = [11 +- sqrt(11^2 - 4(4)(6)] / 2(4)
e^x = [11 +- 5] / 8
e^x = 16/8, 6/8
e^x = 2, 3/4
x = ln2, ln(3/4)
x = 0.693, -0.288

a = 0.693 and b = -0.288

Answer 4

you need to solve for when y=0

as e^x is never 0, you need to solve

0=4e^x + 6e^-x-11

or

-4e^x = 6e^-(x+11)

e^x/(e^-(x+11) = 6/4

e^2x+11 = -6/4

2x +11 = -ln6 - ln4

solve for x and this gives you one

not sure about the other