Prove that for any real number 'a' we have 4a-a^4<=4.?

Question

<= means smaller or equal

Thank you.

Answer 1

You could use calculus.
To prove 4a-a^4<=4
Show that 4 -4a +a^4 >=0 for all a
Minimum value ( or maximum) of 4 -4a +a^4
occurs when d(4 -4a +a^4)da =0
ie -4 +4a^3 =0
4(a^3 -1) =0
ie a=1
Plug in a=1, then 4 -4a +a^4=4-4+1=1
ie you have either a minimum or maximum when a=1
Using the second derivative,
d2(4 -4a +a^4)da2 = 12a^2
when a=1, second derivative =12 >0
therefore you have a minimum at this point.
As the minimum is 1, 4 -4a +a^4 >0 for all a and hence 4a-a^4<=4 for all a.

Answer 2

4a - a^4 - 4 <= 0 Take the derivative w.r.t. a and set equal to 0, we are looking for extrema.

4 - 4a^3 = 0
4 = 4a^3
1 = a^3
1 = a (our extrema)

The second derivative is -12a^2. This everywhere negative (our function is concave downward everywhere) so our extrema is an absolute maximum.

So it is true that 4a - a^4 <= 4 for all a. The biggest it can get is at a = 1 and that gives a value of 4*1 - 1^4 = 4 - 1 = 3

Answer 3

Looking for the maximum of y = 4a - a^4.
4 - 4a^3 = 0
a^3 = 1
a = 1
4*1 - 1^4 = 3 < 4

Answer 4

a = {0, 1}