Question

The graph y=(7x−2)/e^x has a horizontal tangent at x=

Answer 1

This is nitpicky, but it should say `has a horizontal tangent at y =`

the `x=` should be `y=`

Answer 2

anyways, look at a table of values. Let x get larger and larger (10 --> 100 ---> 1000, etc) and see what happens to y

Answer 3

yes the denominator is growing much much faster than the numerator, so overall the fraction gets closer to 0

Answer 4

Actually, I think the \(x=\) is fine, since they use the word "at" rather than "which is"

Answer 5

not sure why you deleted that since you were right

Answer 6

delete what?

Answer 7

That's weird i never deleted a comment.

Answer 8

oh my bad I misread the problem entirely

I though it said asymptote

Answer 9

its okay lol

Answer 10

what you'll need to do is take the derivative and then set it equal to zero, and solve for x

Answer 11

you'll use the quotient rule

Answer 12

We're looking for \(x_0\) such that:\[
\frac{dy}{dx}\Bigg|^{x=x_0} =0
\]

Answer 13

the derivative is e^-x(9-7x)

Answer 14

Then \(9-7x=0\) would give a root.

Answer 15

so does that mean that x=9/7?

Answer 16

h(x) = f(x)/g(x)

let h(x) be set to (7x−2)/(e^x)

so

f(x) = 7x-2
g(x) = e^x

Answer 17

it worked :D thanks man

Answer 18

f(x) = 7x-2
f ' (x) = 7

g(x) = e^x
g ' (x) = e^x

----------------------------

h(x) = f(x)/g(x)

h ' (x) = [f ' (x)*g(x) - f(x)*g ' (x)]/[g(x)]^2

h ' (x) = [7*e^x - (7x-2)*e^x]/[e^x]^2

h ' (x) = [e^x*(7 - 7x+2)]/[e^(2x)]

h ' (x) = [e^x*(9 - 7x)]/[e^(2x)]

h ' (x) = [9 - 7x]/[e^(x)]

so you have the correct derivative

Answer 19

@jim_thompson5910 yeah :D