Identify a counterexample to disprove n3 ≤ 3n2, where n is a real number.
n=-1
n=0
n=4
n=0.5
thanks you
To find a counterexample to disprove the inequality n^3 ≤ 3n^2, you need to find a real number value for 'n' that makes the inequality false. Let's substitute the given values and check.
For n = -1:
n^3 = (-1)^3 = -1
3n^2 = 3(-1)^2 = 3(1) = 3
In this case, n^3 is not less than or equal to 3n^2, -1 > 3, which means the inequality is false.
For n = 0:
n^3 = 0^3 = 0
3n^2 = 3(0)^2 = 3(0) = 0
In this case, n^3 is not less than or equal to 3n^2, 0 > 0, which means the inequality is false.
For n = 4:
n^3 = 4^3 = 64
3n^2 = 3(4)^2 = 3(16) = 48
In this case, n^3 is greater than 3n^2, 64 > 48, which means the inequality is false.
For n = 0.5:
n^3 = (0.5)^3 = 0.125
3n^2 = 3(0.5)^2 = 3(0.25) = 0.75
In this case, n^3 is not less than or equal to 3n^2, 0.125 > 0.75, which means the inequality is false.
By finding these counterexamples, we can conclude that the inequality n^3 ≤ 3n^2 is not universally true for all real numbers.
geez -- did you even bother trying the choices?
clearly, -1^3 is negative, and 3*1^2 is positive, so ✅
Now , in general, if
n^3 ≤ 3n^2
n^2 (n-3) ≤ 0
and since n^2 is never negative, that means
n-3 ≤ 0
n ≤ 3
Did you mean:
n^3 ≤ 3n^2 ?
let n = 5
is 5^3 ≤ 3(5)^2 ?
is 125 ≤ 75 ? , obviously NO
so your hypothesis is bogus, (all you need is one counterexample)