average rate of change?
Problem:
Find the average rate of change of the function f(x)=-3(x^2)-x from x1=5 to x2=6. The formula for average rate of change from a to b is (f(b)-f(a))/b-a
a)-1/6
b)-34
c)-2
d)1/34
e)1/2
help please!
average rate of change?
Problem:
Find the average rate of change of the function f(x)=-3(x^2)-x from x1=5 to x2=6. The formula for average rate of change from a to b is (f(b)-f(a))/b-a
a)-1/6
b)-34
c)-2
d)1/34
e)1/2
help please!
Hello!
All the information you need is given to you. You just need to plug it all in.
You are given your two x values (5 & 6) now, to find f(5) & f(6) just plug them into the function and find your y values.
f(x) = y
When you see f(x) that means that you are solving for y in terms of x
If you are given an x value and a function f(x), you can plug the x value into every single x in the function and it will calculate the y value associated with that x value. So f(5) = -80 See if you can go from there!
Good Luck!
Ok, no problem. Let me try to explain it another way.
The average rate of change of a function is also known as the slope. You know what slope is right?
slope = $\displaystyle \frac{y_2-y_1}{x_2-x_1}$
OR,
slope = $\displaystyle \frac{change-in-y-values}{change-in-x-values}$
OR,
slope = $\displaystyle \frac{f(b)-f(a)}{b-a}$
b & a are just x coordinates somewhere on your function
f(b) & f(a) are the y values associated with those x values.
Since you are given the x values of the coordinates on the function, just plug those into your function to give you the y values.
All that f(x) means is y. f(x) is the function y, evaluated at an x value.
So for example:
$\displaystyle f(x) = 2x^2 - 6x + 3$
Find f(2).
$\displaystyle f(2) = 2(2)^2 - 6(2) + 3$
In calculus,
The average rate of change of a function is the slope of the of the secant line between two points somewhere on the graph. A secant line is a line that goes through the graph of a function in two places. It calculates the average of the slopes of all the tangent line to the function between those two given points.
Hope that clarifies better for you