Question

Which of the following functions grows the slowest?
j(t)=1/4 ln(t^200)
a(t)=t^5/2
i(t)=ln(t^100)
g(t)=3t^2-t
b(t)=t^4-3t+9

Answer 1

in order, slowest to fastest:
log
polynomial
exponential

so, look at the logs
1/4 ln(t^200) = 1/4 * 200 ln t = 50 ln t
ln(t^100) = 100 ln t

so, ...

Answer 2

The growth of a function is its rate of change, which is found by taking its derivative.

Evaluate the derivatives of the given functions and identify the smallest.



j(t) = 1 / 4 ln ( t^200 )

ln ( t^200 ) = 200 ln

so

j(t) = 1 / 4 * 200 ln t

j(t) = 50 ln t

j'(t) = 50 * 1 / t

j'(t) = 50 / t


a(t) = t ^ 5 / 2

a'(t) = 5 / 2 * t ^ ( 5 / 2 - 1 )

a'(t) = 5 / 2 * t ^ ( 5 / 2 - 2 / 2 )

a'(t) = 5 / 2 t ^ ( 3 / 2 )


i(t) = ln ( t^100 )

i'(t) = 1 / t * 100

i'(t) = 100 / t


g(t) = 3 t ^ 2 - t

g'(t) = 3 * 2 * t - 1

g'(t) = 6 t - 1


b(t) = t ^ 4 - 3 t + 9

b'(t) = 4 t ^ 3 - 3


We can rule out the equations:

a(t) = t ^ 5 / 2

g(t) = 3 t ^ 2 - t

and

b(t) = t ^ 4 - 3 t + 9

as their growth is directly related to the variable t.

Meaning that as it gets larger, the functions growth increases.


That leaves equations:

i(t) = ln ( t^100 )

and

j(t) = 1 / 4 ln ( t^200 )

whose growth is inversely related to variable t.

We can see that i'(t) is twice j'(t), so j(t) has the smallest growth.