Suppose #f(x)=x^4-4x^2+2#. What is the average rate of change of the function on the interval [1.9, 2.1]?

Question

Isaiah Allen · Accepted Answer

#16.08# The following yields the average rate of change: #(f(b)-f(a))/(b-a)# #a= 1.9# , #b=2.1# #(((2.1)^4-4(2.1)^2+2)-((1.9)^4-4(1.9)^2+2))/(2.1-1.9)# #((3.8081)-(0.5921))/(0.2)=16.08#

Christopher Allers · Answer

undefined To find the average rate of change of the function $ f(x) = x^4 - 4x^2 + 2 $ on the interval $[1.9, 2.1]$, we first need to evaluate the function at the endpoints of the interval.

$$ f(1.9) = (1.9)^4 - 4(1.9)^2 + 2 $$
$$ = 5.8321 - 14.44 + 2 $$
$$ = -6.6079 $$

$$ f(2.1) = (2.1)^4 - 4(2.1)^2 + 2 $$
$$ = 9.261 - 17.64 + 2 $$
$$ = -6.379 $$

Then, we can use these values to calculate the average rate of change using the formula:

$$ \text{Average rate of change} = \frac{f(2.1) - f(1.9)}{2.1 - 1.9} $$

Substituting the values we found earlier:

$$ \text{Average rate of change} = \frac{-6.379 - (-6.6079)}{2.1 - 1.9} $$
$$ = \frac{-6.379 + 6.6079}{0.2} $$
$$ = \frac{0.2289}{0.2} $$
$$ = 1.1445 $$

Therefore, the average rate of change of the function on the interval $[1.9, 2.1]$ is approximately $1.1445$.