Question

Find the equation of the line tangent to the graph of y=4e^x at x=3.
tangent line y=

Answer 1

any attempt ?

Answer 2

well the tangent line is the derivative right so i got just the same thing 4e^x
im not sure if i am right

Answer 3

you are right

Answer 4

now evaluate it at x=3 so you get 4e^3

Answer 5

thats the slope for any x.
just substitute x=3 and you'll have the slope of the curve at x=3
for the equation, you already know the slope, you'd just need a point . You already know that point, don;t you ?

Answer 6

well i plugged in that for my answer and i got it incorrect so i was a bit confuse

Answer 7

Can you find the derivative of y=4(e^x)?

Answer 8

Also, are you looking for point slope form, or slope intercept form?

Answer 9

well i thought that the derivative was just the 4e^x but im not sure and it just tells me to find the tangent equation

Answer 10

well by 1st task in this question is to for the y value of the point where x = 3

so

\[y = 4e^3\]

so the point is

\[(3, 4e^3)\]

next find the 1st derivative, which you have done

\[f'(x) = 4e^x\]

find the slope when x = 3

\[m = 4e^3\]

so using the equation of the line

y = mx + b

substitute and you get

\[4e^3 = 4e^3(3) + b\]

all you need to do is solve for b, the y- intercept

\[4e^3 = 12e^3 + b\]

then your equation will be

\[y = 4e^3x + b\]

hope this helps.

Answer 11

so i got 4e^3x+4 is that correct

Answer 12

Why +4?

Answer 13

\(4e^3 = 12e^3 + b \implies 4e^3-12e^3 = b\)

Answer 14

ok first calculate derivative 0f Y
that is = 4e^x
that is the slope of tangent
at x=3 slpoe =4e^3
tangent is passing thru x=3
so find y point
y=4e^x
at x=3 y = 4e^3
(3,4e^3)

so the eqation of line Y=mX+c
put x, y ,slope in it
4e^3=4e^3*3+c
c=-8e^3
y=4e^3x-8e^3