Rates of changeRates of change look at how two things change relative toone another; phrased at functions it is the change of theoutput relative to the change of the input (∆y/∆x). If wethink to our basic flat object of interest (lines) then the rateof change is the slope of the line.slope =riserun=y2− y1x2− x1=∆y∆xyxb∆x = x2− x1∆y = y2− y1(x1, y1)(x2, y2)For functions that are not line there is not “one” rate ofchange so we have to change our approach. The first thingthat we look at is the average rate of change of the functionfrom x = a to x = b. This is the slope of the secant linepassing througha, f(a)andb, f(b).abf(b)f(a)secant liney = f(x)One way to think about the average rate of change findwhat mahoganyconstant rate of change produces the sameeffect to go froma, f(a)tob, f(b). [It is important tonote that when computing the average rate of change wedon’t know what happens to the functions between theendpoints; it might not even be defined!]A different rate of change is the instantaneous rate of changewhich looks at the rate of change of the function at aparticular instant. Intuitively when we zoom in on a pointof the function, if it looks like a line then the slope of thatline should be the instantaneous rate of change. We callthis line the tangent line (since it “touches” at one point).The problem of computing the slope of the tangent line isthat there is only one point of contact. So now we have theimportant idea: we can start with secant lines and thenmove the points closer together to form a series of lines(shown dotted in next figure) that get closer to the tangentline (shown solid in the next figure).In other words as our two points get “closer and closer”the average rate of change approaches the instantaneousrate of change. Of course, now we have to figure out howwe get “closer and closer”; which we pick up next time.xya ←− b ←− b ←− bProblems1. Find the average rate of change for y = 5x + 219 fromx = π5to x = e17.2. Find the average rate of change fory = ex+ x2+ tan(2x) from x = 0 to x = 5.3. Given that f(1) = 3, f(9) = 11, and that the averagerate of change of f(x) from x = 1 to x = 4 is twice theaverage rate of change of f(x) from x = 4 to x = 9,determine f(4).4. Looking over your lab notes of an experiment whereyou are looking at the height of a certain object inmeters at a given time in seconds you note that yourecorded the following observations.• At 1 second the height is 3 meters.• At 8 seconds the height is 1 meter.You recall that you made one more observation buthave lost the data in your notes. You do recall thefollowing:• From the first observation to the missingobservation the average rate of change was equalto 1.• From the second observation to the missingobservation the average rate of change was equalto −2.Determine at what time in seconds the missingobservation took place. (For added fun; do this usingas little algebra as possible.)5. If the average rate of change of f(x) from x = 1 tox = 4 is 1 and the average rate of change of xf(x) fromx = 1 to x = 4 is 6, find the average rate of change ofx2f(x) from x = 1 to x = 4.6. Find the tangent line to the parabola y = 3x2− 7x + 3at x = 2.7. If the average rate of change of f(x) from x=1 to x=3is 2 and the average rate of change off(x)2fromx=1 to x=3 is 20, determine the average rate of changeoff(x)3from x=1 to