Question #264736

Determine for which positive integer values of n, 3n^3+2≤n^4 and prove your claim by mathematical induction.


I was able to determine n must be greater or equal to 4. But I can't seem to get through the inductive step.


1
Expert's answer
2021-11-14T16:26:06-0500

Let f(x)=3x3+2x4,x>0f(x)=3x^3+2-x^4, x>0


f(x)=9x24x3f'(x)=9x^2-4x^3

Find the critical number(s)


f(x)=0=>9x24x3=0f'(x)=0=>9x^2-4x^3=0

x2(94x)=0x^2(9-4x)=0

x1=0,x2=2.25x_1=0, x_2=2.25

If x<2.25,f(x)>0,f(x)x<2.25, f'(x)>0, f(x) increases.

If x>2.25,f(x)<0,f(x)x>2.25, f'(x)<0, f(x) decreases.



f(1)=3+21=4>0f(1)=3+2-1=4>0

f(2)=24+216=10>0f(2)=24+2-16=10>0

f(3)=81+281=2>0f(3)=81+2-81=2>0

Therefore 3n3+2n43n^3+2\leq n^4 for n4n\geq4.


Let P(n)P(n) be the proposition that 3n3+2n4,n4.3n^3+2\leq n^4, n\geq4.

Basis Step

P(4)P(4) is true, because 3(4)3+2=194<256=(4)4.3(4)^3+2=194<256=(4)^4.

Inductive Step

We assume that 3k3+2k4.3k^3+2\leq k^4. Under this assumption, it must be shown that P(k+1)P(k + 1) is true, namely, that


3(k+1)3+2(k+1)43(k+1)^3+2\leq (k+1)^4

We have


3(k+1)3+2=3k3+9k2+9k+3+23(k+1)^3+2=3k^3+9k^2+9k+3+2

k4+(4k)k2+(6k)k+(4k+1)=(k+1)4,k4\leq k^4+(4k)k^2+(6k)k+(4k+1)=(k+1)^4, k\geq4

Hence P(k+1)P(k + 1) is true under the assumption that P(k)P(k) is true. This completes the inductive step.

We have completed the basis step and the inductive step, so by mathematical induction we know that P(n)P(n) is true for integers n4n\geq4. That is, we have proven that


3n3+2n4,n4.3n^3+2\leq n^4, n\geq4.

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