Brain Teaser 6: Differentiate y = Ae^2x + Be^-x, where x ≥ 0,…

A curve has equation y = Ae^2x + Be^-x, where x ≥ 0. At the point where x = 0, y = 50 and dy/dx = -20.

i)             Show that A = 10 and find the value of B.                                             (5)

ii)           Using the values of A and B found in part (i), find the coordinates of the stationary point of the curve.                                                                 (4)

{My additional question: Express the exact values of the coordinates of this stationary point as     (C ln 2, D(2)^(2/3) ) and show that C = 1/3 and D = 30)

iii)          Determine the nature of the stationary point, giving a reason for your answer.     (2)

-------------------------------------------------------------------------------------------------------

In an earlier post, I have shown that e^{ln 2} = 2, 10^{lg 3} = 3 and generally b^log_b x = x. My additional question in part (ii) may require the use of this knowledge in order to get the exact value of the y-coordinate of the stationary point mentioned therein.

(Background to this Q: In the course of preparing a student to sit for her IGCSE Cambridge Add-Maths 0606 next Tuesday (10/06/2014), I came across the above Q as one of the past-year questions on "Calculus". After teaching her how to get the answers, I additionally asked her to get the exact values of the coordinates of the stationary point instead of just (0.231, 47.6). Thus, I have added an additional question to part (ii). :) )

-------------------------------------------------------------------------------------------------------

Answers:

i)                    A = 10; B = 40

ii)                   Stationary point = (0.231, 47.6) [or, ((1/3)ln 2, 30(2)^(2/3)); C = 1/3: D=30]

iii)                 d²y/dx² = 40e^2x + 40e^-x = +ve for all values of x – thus, minimum point.

Working leading to the answers will be posted in a week or so (please see below on "My Proposed Solutions"). You may post yours by way of comments if this question is of interest to you too. J

(Posted on 3/06/2014 from PJ SS2 Starbucks)

--------------------------------------------------------------------------------------------------------

My Proposed Solutions:

(i)             Given :    y = Ae^2x + Be^-x, where x ≥ 0

              At x = 0, y = 50 and dy/dx = -20

At x = 0, y = 50, therefore:          50 = Ae^o + Be^o

                                                   50 = A + B          ~~~ Eqn (1)

At x = 0, dy/dx = -20, therefore:  dy/dx = 2Ae^2x – Be^-x

                                                   -20  =  2Ae²⁽⁰⁾ – Be^-(0)

                                                   -20 = 2A – B      ~~~ Eqn (2)

Eqn (1) + Eqn (2):                       3A = 30,          hence, A = 10 (Shown)

Use Eqn (1):                               50 = 10 + B,    hence, B = 40 (Answer)

(ii)           Since,                 A = 10 and b = 40

therefore:            dy/dx = 2Ae^2x – Be^-x = 20e^2x – 40e^-x

At stationary pt, dy/dx = 0

therefore:            0 = 20e^2x – 40e^-x

x (e^x):                  0 = 20e^2x(e^x) – 40e^-x(e^x)

                           0 = 20e^3x – 40e^o

                           0 = 20e^3x – 40

                           e^3x = 40/20 = 2

ln both sides:       3x = ln 2.         Therefore,        x = (1/3)(ln 2) ~~ exact value

                                                                           x = 0.231 (3 sf) ~~ (Answer)

Put x = (1/3)(ln 2) into:   y = 10e^2x + 40e^-x

therefore,                        y = 10e^{2(1/3)(ln 2)} + 40e^{-(1/3)(ln 2)}

                                       y = 10e^{(ln 2^(2/3))} + 40e ^{(ln 2^(-1/3))}

[Use ln e^x = x:                 y = 10(2^(2/3)) + 40(2^(-1/3))]

All in terms of 2^(2/3):        y = 10(2^(2/3)) + 40(2^(-1/3))(2/2)

                                       y = 10(2^(2/3)) + [40(2^(-1/3))(2)]÷2

                                       y = 10(2^(2/3)) + [40(2^(1-(1/3)))] ÷2

                                       y = 10(2^(2/3)) + 20(2^(2/3)) = 30(2^(2/3))      ~~ exact value

                                       y = 47.6 (to 3 sf)               ~~ (Answer)                                       

(iii)          d²y/dx² = 40e^2x + 40e^-x = +ve value for all x ≥ 0

So, stationary pt is minimum.                                ~~ (Answer)

Please drop your comments if you have any view to express on the above. Have fun!

(Posted on 18/06/2014 from PJ SS2 Starbucks)