Draw a sketch of a curve y=e^(-2x) -3x

[Unusualskill]

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 stop at part c.

 

[Quaid]

i stop at part c.

Please post your work on parts (a) and (b).

Cheers ~ Mark :cool:

 

[Unusualskill]

Please post your work on parts (a) and (b).

Cheers ~ Mark :cool:

(a) At A(a,0)

0=e^(-2a) -3a

(b) dy/dx=-2e^(-2x) -3
=-2e^(-2a)-3

equation of tangent,
y-0= (-2e^-2ax-3)(x-a)
y=-2xe^-2a - 3x +2ae^-2a +3a
when x=0,
y=2ae^-2a +3a (shown)

 

[Unusualskill]

the trick is to remember that $\displaystyle e^{-2a}=3a$

i don understand for part c , where is the < 1 comes from. obviously i can see 6a^2+3a .

 

[stapel]

i don understand for part c , where is the < 1 comes from. obviously i can see 6a^2+3a .

What do you mean by "obviously i can see" the one side of the inequality, but not the other? What have you tried? How far have you gotten? Where are you stuck?

Please show your work completely. Thank you!