Reverse Differentiating e (1 Viewer)

[Finx]

Find the equation of the curve that has f''(x) = 12e^(2x) and a stationary point at (0,3).

If I remember correctly, stationary points are where the gradient = 0.

y'' = 12e^(2x)
y' = 6e^2x + c

Is this right so far? I'm just stuck on where to go next. Thanks in advance!

 

[lolokay]

you have the expression for the gradient, and you know the value of x (and corresponding y value) for which the gradient is zero. given this you can evaluate the constant

 

[Finx]

y' = 6e^(2x) + c

I know the gradient is zero, so;

0 = 6e^(2x) + c

I don't know what x or c is..

Stationary point at (0,3) means [y' = 0] and the curve passes through [0,3]. Am I missing something?

 

[lolokay]

if there is a "Stationary point at (0,3)" then what is x?

 

[shinn]

 

[Finx]

Ah, so stationary point means y' = 0 and x = 0?

Edit; the answer is 3e^2x - 6x.

 

[lolokay]

yep