SOLUTION: Evaluate the functions for the values of x. Given as 1,2,4,8,16. Describe the difference in the rate at which each function changes with increasing values of x. f(x)=3x+2 f(x)=x^

Algebra ->  Algebra  -> Absolute-value -> SOLUTION: Evaluate the functions for the values of x. Given as 1,2,4,8,16. Describe the difference in the rate at which each function changes with increasing values of x. f(x)=3x+2 f(x)=x^      Log On

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Question 91084: Evaluate the functions for the values of x. Given as 1,2,4,8,16. Describe the difference in the rate at which each function changes with increasing values of x.
f(x)=3x+2
f(x)=x^2+5x+6
f(x)=x^3+3x^2+2x+1
f(x)=e^x
f(x)=logx
This is for my daughter and I am helping with homework and I need to know I am right or how to explain it

Answer by mathispowerful(115) About Me  (Show Source):
You can put this solution on YOUR website!
I would skip evaluation part, that's tedious calculation.
for the second part, observe the following graphs
f(x)=3x+2 --- brown
f(x)=x^2+5x+6 -- green
f(x)=x^3+3x^2+2x+1 -- blue
f(x)=e^x --- purple
f(x)=logx --- close to x axis
from the slopes you can see the rate of change, the bigger the slope the bigger the rate of change.

graph%28300%2C600%2C-5%2C10%2C-10%2C80%2C3x%2B2%2Cx%5E2%2B5x%2B6%2Cx%5E3%2B3x%5E2%2B2x%2B1%2C2.71828%5Ex%2C+%28ln%28x%29%29%2Fln%2810%29%29
It is hard to tell which rate is bigger between the blue one and purple one, the following graph is the graphs of derivatives of f(x)=x^3+3x^2+2x+1 and
f(x)=e^x , the green one is for f(x)=e^x, and green one eventually goes above f(x)=x^3+3x^2+2x+1, that means e^x has bigger rate of change than x^3+3x^2+2x+1
after a certain value of x.

graph%28200%2C300%2C+-2%2C+6%2C+-5%2C+50%2C+3%2Ax%5E2%2C+2.71828%5Ex%29
the conclusion:
the increase rate is in the following order, from smallest to biggest
f(x)=logx
f(x)=3x+2
f(x)=x^2+5x+6
f(x)=x^3+3x^2+2x+1
f(x)=e^x

hope this helps you and your daughter.