Let f(x)=3x3+2−x4,x>0
f′(x)=9x2−4x3 Find the critical number(s)
f′(x)=0=>9x2−4x3=0
x2(9−4x)=0
x1=0,x2=2.25 If x<2.25,f′(x)>0,f(x) increases.
If x>2.25,f′(x)<0,f(x) decreases.
f(1)=3+2−1=4>0
f(2)=24+2−16=10>0
f(3)=81+2−81=2>0 Therefore 3n3+2≤n4 for n≥4.
Let P(n) be the proposition that 3n3+2≤n4,n≥4.
Basis Step
P(4) is true, because 3(4)3+2=194<256=(4)4.
Inductive Step
We assume that 3k3+2≤k4. Under this assumption, it must be shown that P(k+1) is true, namely, that
3(k+1)3+2≤(k+1)4 We have
3(k+1)3+2=3k3+9k2+9k+3+2
≤k4+(4k)k2+(6k)k+(4k+1)=(k+1)4,k≥4Hence P(k+1) is true under the assumption that 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) is true for integers n≥4. That is, we have proven that
3n3+2≤n4,n≥4.
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