Expon and log functions question. (1 Viewer)

[RHINO7]

Please help needed-

i) Find the equation of the normal to the curve y=e^x at the point where x=3, in exact form.




ii)Find the stationary point on the curve y=xe^x and determine its nature.




iii) Find the second derivative of (e^2x +1)^7




iv) Find the equation of the tangent to the curve y=e^2x -3x at the point (0,1)



Thank you

 

[jb_nc]

lol, no joke man you should probably drop down to general

 

[RHINO7]

hey no joke MAN, i've done the questions i wanted to see if i got the working out correct. Dick.

 

[S1M0]

This may be year 12 stuff, but i'll just guess.

1) y = e^3??

 

[jb_nc]

This may be year 12 stuff, but i'll just guess.

1) y = e^3??

It says the equation and normal to the curve. So,

y = e^x
dy/dx = e^x

at x = 3, dy/dx = e^3 but we know that iff m1*m2 = -1 then it is the normal gradient is -(1/e^3).

The point (y1) is e^3.

y - y1 = m(x-x1)
y - e^3 = (-1/e^3)[x - 3]
-e^3(y - e^3) = x - 3
-e^3y - e^6 + 3 = x

(ii) d(xe^x)/dx = e^x + xe^x
Then, e^x(x + 1)
Set dy/dx to 0
e^x(x + 1) = 0
e^x cannot = 0, x = -1 so there is one stationary point.

d^2(xe^x)/dx^2 = e^x + e^x + xe^x = 2e^x + xe^x
Sub in -1 d^2x/dx^2 > 0 so -1 is a MINIMUM point

Point is [-1, -(1/e)]. Domain is R, Range is [-1/e, infinity)

(iii) y=e^2x -3x
dy/dx = 2e^2x - 3
at x = 0, dy/dx = -1
y - y1 = x(x - x1)
y -1 = -(x - 0)
x = -y + 1

There

 

[ellen.louise]

This may be year 12 stuff, but i'll just guess.

1) y = e^3??

bad decision. it's a no brainer but you got it wrong. go back to your own forum, junior!

 

[darkliight]

Please help needed-

iv) Find the equation of the tangent to the curve y=e^2x -3x at the point (0,1)

Thank you

We've got this curve out in our 2D plane right? We don't care what it looks like, but it makes sense that at any point along that curve we could draw a line that just touches the curve at that point, right? Something like

Now, the equation of a line is given by y - y_1 = m(x - x_1) right? So for our tangent line, all we need to know is a point (x_1, y_1) our line passes through and its slope (or rate of change), m, so we can plug it into that equation. We're already given the point, we just need to know its slope. For a tangent line, the rate of change we want is the same as the rate of change of our curve at that point. We know the rate of change of y = e^2x -3x is given by dy/dx = 2e^2x -3 right? In particular, we want to know the rate of change at the point (0,1), so plugging that into our derivative we get m = 2e^(2*0) - 3 = 2 - 3 = -1.

We know everything now, so plugging it into the equation of a line we wrote down before, we have y - 1 = -1(x - 0) ==> y = 1 - x.

I was extra bored so here is what it looks like:

Also, whats with the hostility in this thread?