Question

Use the following graph of the function f(x) = 3x4 − x3 + 3x2 + x − 3 to answer this question: graph of 3 x to the fourth, minus x cubed, plus 3 x squared, plus x minus 3. What is the average rate of change from x = 0 to x = 1?

Answer 1

Answer:

Average rat of change of function is 6 from x = 0 to x = 1.

Step-by-step explanation:

We are given a function:

$f(x) = 3x^4 - x^3 + 3x^2 + x - 3$

The attached image shows the graph for the given function.

Average rate of change of function =

$\text{Average rate of change} = \displaystyle\frac{\delta y}{\delta x} = \frac{f(b)-f(a)}{b-a}$

Now we will evaluate f(1) and f(0).

$f(x) = 3x^4 - x^3 + 3x^2 + x - 3\\f(1) = 3(1)^4 - (1)^3 + 3(1)^2 + (1) - 3 = 3\\f(0) = 3(0)^4 - (0)^3 + 3(0)^2 + (0) - 3 = -3$

Putting the values, we get,

$\text{Average rate of change} = \displaystyle\frac{f(1)-f(0)}{1-0}\\\\= \frac{3-(-3)}{1-0} = \frac{6}{1} = 6$

Thus, the average rat of change of function is 6 from x = 0 to x = 1.

Answer 2

For this case we have the following function:
 $f (x) = 3x ^ 4 - x ^ 3 + 3x ^ 2 + x - 3$
 By definition we have that the average rate of change is given by:
 $AVR = \frac{f(x2)-f(x1)}{x2-x1}$
 We evaluate the function for the given interval:
 For x = 0:
 $f (0) = 3 (0) ^ 4 - (0) ^ 3 + 3 (0) ^ 2 + (0) - 3 f (0) = - 3$
 For x = 1:
 $f (1) = 3 (1) ^ 4 - (1) ^ 3 + 3 (1) ^ 2 + (1) - 3 f (1) = 3$
 Substituting values we have:
 $AVR = \frac{3-(-3)}{1-0}$
 Rewriting we have:
 $AVR = \frac{3+3}{1-0}$
 $AVR = \frac{6}{1}$
 $AVR =6$
 Answer:
 the average rate of change from x = 0 to x = 1 is:
 $AVR =6$