i'm not sure how to do this. can someone help, please? thanks!
consider the function f(x0 = x^2 + 2x on the interval [-2, 2]
a. draw a sketch of the graph of f(x). find the average rate of change on the interval [-2, 2] and sketch this secant line.
b. find an expression that calculates the instantaneous rate of change of f(x) at any given point on the curve. find the coordinates on the graph of f(x) where the instantaneous rate of change equals the average rate of change.
c. write an equation for the tangent line at the point found in part b. sketch the tangent line on the same graph as part a. what is the relationship between the two lines on the graph?
thanks for the help in advance, i'm just really confused.
a.)Sorry but I cant post the link to the graph so you gotta do that part with a calculator
To find average rate of change you need your two points with the given X values [-2,2]. Those points end up being (-2,0) and (2,8). Find the slope:
(8-0)/(2-(-2))=8/4=2
b.) Take derivative of your equation:
f'=2x+2
Since you wanna know where instantaneous rate of change and average rate of change are equal, set your derivative equal to your avg.:
2=2x+2
x=0
Plug your new x value into the original equation to get your coordinates (0,0)
c.)y=2x
They should be parallel.
p.s. I don't know if you need this, but the eqn. of your secant line is y=2x+4
Sure, I can help you with that! Let's go step by step:
a. To draw a sketch of the graph of f(x), we can start by plotting some points on the graph. Plug in some values from the interval [-2, 2] into the function f(x) = x^2 + 2x, such as -2, -1, 0, 1, and 2, and calculate the corresponding y-values. Then plot these points on a graph. Connect the points to get a smooth curve. Label the x and y axes as appropriate.
To find the average rate of change on the interval [-2, 2], we can use the formula:
Average rate of change = (f(2) - f(-2)) / (2 - (-2))
Compute the values of f(2) and f(-2) by plugging in 2 and -2, respectively, into the function f(x). Then substitute these values into the formula to calculate the average rate of change.
The secant line is a straight line that connects two points on a curve. On your graph, draw a line that passes through the two points (-2, f(-2)) and (2, f(2)). This line represents the average rate of change on the interval [-2, 2].
b. The instantaneous rate of change of a function at any given point on the curve is represented by the derivative of the function. To find an expression that calculates the instantaneous rate of change of f(x), we need to take the derivative of the function f(x) = x^2 + 2x.
Derivative of f(x) = d/dx (x^2 + 2x)
Differentiate each term in the function separately using the power rule:
Derivative of x^2 = 2x (since the exponent 2 comes down as a coefficient)
Derivative of 2x = 2 (since the derivative of a constant times x is the constant).
Combine the derivative terms to get the expression for the instantaneous rate of change of f(x).
To find the coordinates on the graph of f(x) where the instantaneous rate of change equals the average rate of change, we need to solve the equation:
Derivative of f(x) = Average rate of change
Substitute the expression for the derivative obtained earlier and the value of the average rate of change calculated in part a into the equation. Solve for x to find the x-coordinate(s). Then substitute the x-coordinate(s) back into the function f(x) = x^2 + 2x to find the corresponding y-coordinate(s).
c. To write an equation for the tangent line at the point found in part b, we need the slope of the tangent line and the coordinates of the point. The slope of the tangent line is equal to the instantaneous rate of change of the function at that point.
Using the expression for the instantaneous rate of change obtained in part b, substitute the x-coordinate(s) found earlier to calculate the slope of the tangent line. Then use the point-slope form of a line (y - y1 = m(x - x1)), where (x1, y1) is the point found earlier and m is the slope, to write the equation of the tangent line.
Sketch the tangent line on the same graph as part a. The tangent line will touch the curve of f(x) at the point found earlier and have a slope equal to the instantaneous rate of change at that point.
The relationship between the secant line from part a and the tangent line from part c is that the secant line represents the average rate of change on the interval [-2, 2], while the tangent line represents the instantaneous rate of change at a specific point. The tangent line is more closely aligned with the curve at that specific point than the secant line, which spans the entire interval.