Log Questions (1 Viewer)

[Mc_Meaney]

I have two questions:

1) find f'(x) if f(x) = x³.In(x+1)

According to the answers, the answer is be 3x²+ x³/(x+1) - i just dont get to that :burn:

2) Find any stat points on y= x²e^2x

I have no idea how to do this question

Any help would be appreciated

TY in advanced.

 

[Mc_Meaney]

bhump

 

[SoulSearcher]

For the first one, if I read the question correctly and it's x³ln(x+1), then its derivative should be 3x²ln(x+1) + x³/(x+1).

For the second one, y=x²e^2x
y' = 2xe^2x + 2x²e^2x
= 2xe^2x(1+x)
Stationary points occur when y' = 0
0 = 2xe^2x(1+x)
Therefore either
2xe^2x = 0 or 1+x = 0
Therefore x = 0 and x = -1
When x = 0, y = 0
x = -1, y = e^-2
Therefore stationary points are (0,0) and (1,e^-2)

 

[Mc_Meaney]

For the first one, if I read the question correctly and it's x³ln(x+1), then its derivative should be 3x²ln(x+1) + x³/(x+1).

For the second one, y=x²e^2x
y' = 2xe^2x + 2x²e^2x
= 2xe^2x(1+x)
Stationary points occur when y' = 0
0 = 2xe^2x(1+x)
Therefore either
2xe^2x = 0 or 1+x = 0
Therefore x = 0 and x = -1
When x = 0, y = 0
x = -1, y = e^-2
Therefore stationary points are (0,0) and (1,e^-2)

Thanks SS, thats what i thought the answer to one was as for two, youre a legend.

 

[Mc_Meaney]

New Question for the math geniuses among us-

1) Evaluate ^0-1∫e^3t+4dt (3 dp)

This is my working, Im not sure it if is correct -

[1/3.e^3t+4+t²/2] 0-1

And yeh Im stuck from there lol. any help would be much appreciated :burn:

 

[alcalder]

Evaluate ^0-1∫e^3t+4dt (3 dp)

Well,

∫e^atdt = (1/a)e^at + C

And if you have, (since ^b is just a constant)

∫e^at+bdt = e^b∫e^atdt

= (e^b/a)e^at + C OR

= 1/a(e^at+b) + C

^0-1∫e^3t+4dt = [(1/3)e^3t+4]^0-1
=[e⁴/3 - e¹/3]

= 17.293

It's been a very long time since I've done that. But you do not need to integrate the power of e as well, in this case.