- Use the functions f(x) = 4x + 3 and g(x) = 2x2 + 2 to answer parts (a)-(c). a. Which function has a greater rate of change from x = 0 to x = 1?
b. Which function has a greater rate of change from x = 2 and x = 3?
C. Does g(x) ever have a greater rate of change than f(x)? Explain.

9514 1404 393

  a) f(x)

  b) g(x)

  c) for all values of x > 1, g(x) has a greater rate of change

Step-by-step explanation:

a) The rate of change of f(x) is the x-coefficient: 4. The average rate of change of g(x) on an interval can be found by dividing the change by the interval width. Here, the width of the interval of interest is 1-0 = 1, so the average rate of change is g(1) -g(0) = 4-2 = 2.

On the interval [0, 1], f(x) has the greater rate of change.

__

b) The interval [2, 3] also has a width of 1, so the rate of change of g(x) on that interval is g(3) -g(2) = 20 -10 = 10. This value is greater than 4, so ...

on the interval [2, 3], g(x) has the greater rate of change.

__

c) Part (b) demonstrates that g(x) will have a greater rate of change on some intervals. In fact, for any interval whose center is greater than x=1, g(x) will have a greater rate of change than f(x).

_____

In calculus terms, the rate of change of g(x) is g'(x) = 4x. This will be greater than 4 for any x > 1. The graph shows the rate of change is 4 at x=1, and is higher for x > 1. The average rate of change will be greater than 4 on any interval whose center is greater than x=1.

We can figure the average rate of change on the interval [a, b] as ...

  m = (g(b) -g(a))/(b -a)

  m = ((2b² +2) -(2a² +2))/(b -a) = 2(b² -a²)/(b-a) = 2(b+a)

For the average rate of change to exceed 4, the sum of the ends of the interval must exceed 2, which is to say the midpoint must exceed 1.

250

step-by-step explanation:

this linear equation would have just one solution.

step-by-step explanation:

the reason why this would have one solution is that it is an expression with just one variable. since that variable is not raised to a power, it will always have just one solution.