Pls help!

[cheesecake]

Can folks please help me with another question? Thanks

Draw a sketch of a curve y=e^(-2x) -3x . The curve crosses x-axis at A(a,0) and the y-axis at B(0,1). O is the origin

(a) write down an equation satisfied by a.
(b) Show that the tangent at A meets the y-axis at the point whose y-coordinate is 2ae^(-2a) +3a
(c) show that d2y/dx2>0 and using the results from parts(a) and b ,deduce that 6a^2+3a<1
(d) find ,in terms of a,the area of the region bounded by the curve and the line segments OA and OB
(e) By comparing this area with the area of the triangle OAB, show that 3a^2+4a>1 Hence show that (√7)/3 - 2/3 < a < (√33)/12 - 1/4

I have difficulty with c and e

 

[Dr.Peterson]

Draw a sketch of a curve y=e^(-2x) -3x . The curve crosses x-axis at A(a,0) and the y-axis at B(0,1). O is the origin

(a) write down an equation satisfied by a.
(b) Show that the tangent at A meets the y-axis at the point whose y-coordinate is 2ae^(-2a) +3a
(c) show that d2y/dx2>0 and using the results from parts(a) and b ,deduce that 6a^2+3a<1
(d) find ,in terms of a,the area of the region bounded by the curve and the line segments OA and OB
(e) By comparing this area with the area of the triangle OAB, show that 3a^2+4a>1 Hence show that (√7)/3 - 2/3 < a < (√33)/12 - 1/4

I have difficulty with c and e

Please show what you have been able to do (a, b, d, and whatever you can in c and e), so we can see what help you need.

 

[cheesecake]

a. When x= a
y=e^-2x -3a

b. dy/dx= -2e^-2x -3
When x=a, dy/dx = -2e^-2a -3
Equation of tangent at (a,0)
y=(-2e^-2a-3) (x-a)
When x=0, y= 2ae^-2a +3a

d.integrate e^-2x -3x from 0 to a
= [-1/2e^-2x -3x^2/2] from 0 toa
=(-1/2e^-2a -3x²/2) - (-1/2 e^0 -0)
=1/2 - 3/2 a² -1/2 e^-2a

 

[Dr.Peterson]

a. When x= a
y=e^-2x -3a

But what is y, when x = a? And why did you replace only one x with a?

b. dy/dx= -2e^-2x -3
When x=a, dy/dx = -2e^-2a -3
Equation of tangent at (a,0)
y=(-2e^-2a-3) (x-a)
When x=0, y= 2ae^-2a +3a

Good.

d.integrate e^-2x -3x from 0 to a
= [-1/2e^-2x -3x^2/2] from 0 toa
=(-1/2e^-2a -3x²/2) - (-1/2 e^0 -0)
=1/2 - 3/2 a² -1/2 e^-2a

When you get a useful answer for (a), you can combine that with this, and that will help in (e).

For (c), did you at least try finding the second derivative?

 

[cheesecake]

a) y=e^-2a-3a

c) d2y/dx2= 4e^-2x
when x=0, d2y/dx2=4>0
e^-2a-3a=2ae^-2a+3a
2ae^-2a-e^-2a+6a=0

i dont now how e^-2a=3a
but if this is so, then i can get
6a^2+3a

 

[cheesecake]

a) y=e^-2a-3a

c) d2y/dx2= 4e^-2x
when x=0, d2y/dx2=4>0
e^-2a-3a=2ae^-2a+3a
2ae^-2a-e^-2a+6a=0

i dont now how e^-2a=3a
but if this is so, then i can get
6a^2+3a

I was wrong in my previous comment
this should be the woring for part c

c. d2y/dx2=4e^-2x
when x=0, d2y/dx2=4>0

e^-2a-3a=0
e^-2a=3a
then, 2ae^-2a +3a = 2a(3a) +3a
=6a^2 +3a

then how to prove that it s > than 1?
from the graph?

 

[Dr.Peterson]

e^-2a-3a=0

THIS is the answer for part (a)!

c. d2y/dx2=4e^-2x
when x=0, d2y/dx2=4>0

Not just when x=0! You were told to show that it is always positive, which is clearly true if you know how the exponential function works.

e^-2a-3a=0
e^-2a=3a
then, 2ae^-2a +3a = 2a(3a) +3a
=6a^2 +3a

You're saying that the tangent's y-intercept is 6a^2 +3a, right? That's correct.

then how to prove that it s > than 1?
from the graph?

You mean < 1, right?

If the curve is concave up, how will the y-intercepts of the tangent and the curve itself be related? Sketch the part of the curve between the intercepts if you need help thinking about this, though that's not an integral part of the proof.