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A Level Maths Differentiation Question

Hi TSR,

I’m stuck on this differentiation question and any help will be appreciated.

Find the range of the function:

g(x) = (x^2)(1-x)(e^-2x)

Thanks a lot
Reply 1
Avatar for mqb2766
mqb2766
21
Are there any limitations on x?

Few observations
x^2 is always non-negative & large when x is large (quadratic)
e^-2x is always non-negative & large when x is -large and small when x is +large (exponential)
1-x is both positive x<1 & negative x>1 (linear)

When x is -large the negative exponential will dominate g() and the response will be +large
When x = 0, g(0) = 0 and g(1) = 0
When x is +large , g() will be negative as two terms are positive and the linear term is negative.
There is an obvious stationary point at 0 but possibly another two bracketing x=1.
Differentiate and see if you can find them

This should give you the range?
Reply 2
Avatar for Yodalam
Yodalam
OP
14
Thanks, the domain is x is greater or equal to 0
Reply 3
Avatar for mqb2766
mqb2766
21
Just ignore the part of the discussion which relates to negative x then.
I think you have a couple of proper stationary points which will determine the min/max values & hence the range of g() which should pop out of the differentiation?
Reply 4
Avatar for Yodalam
Yodalam
OP
14
Yes, on the given graph there are 2 stationary points- one max and one minimum. I have used differentiation to find both the points but I struggled with that. Any help?

Thanks
g(x) = (x2)(1 - x)(e-2x)
= (x2 - x3)(e-2x)

Differentiating by using the product rule with u = (x2 - x3) and v = (e-2x) ,

g'(x) = -2(x2 - x3)(e-2x) + (2x - 3x2)(e-2x)

At the turning points, g'(x) = 0.

-2(x2 - x3)(e-2x) + (2x - 3x2)(e-2x) = 0

Dividing both sides by (e-2x),


-2(x2 - x3) + (2x - 3x2) = 0

Factorising,

x (-2(x - x2) + (2 - 3x)) = 0
x ( 2x2 - 2x + 2 - 3x) = 0
x ( 2x2 - 5x + 2) = 0
x (2x - 1)(x - 2) = 0

Turning points are at x = 0, x = 1/2, x = 2


Here is the graph: https://www.desmos.com/calculator/8udjojtcin
(edited 5 years ago)
Reply 6
Avatar for mqb2766
mqb2766
21
Original post by Yodalam
Yes, on the given graph there are 2 stationary points- one max and one minimum. I have used differentiation to find both the points but I struggled with that. Any help?

Thanks


What are you struggling with?
g'(x) = quadratic*x*exponential
For g'(x) = 0, you just have to find the roots of the quadratic?
Reply 7
Avatar for Yodalam
Yodalam
OP
14
I have completed it now!!!!

Thanks for your help!! :smile:

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