For the graphed function f(x) = (2)x + 2 + 1, calculate the average rate of change from x = 0 to x = 2. @ranga
@shamil98
@kc_kennylau
What is (2)x + 2 +1? \[2x + 2 + 1\] or \[x^2 + 2x + 1\] this looks to me to be an exponential function of the form \[2^{x+2} + 1\]
ah nevermind... I answered my own question
lol it took you soo long to type that? yes this is a exponential function.
you need to know 2 things here to do this derivative: the chain rule and that: \[dA^x = ln(A)A^xdx\] \[d(2^{x+2} + 1) = ln(2)2^{x+2}dx\]
ahh just realized you don't want the calculus version of this.
So how do I apply these rules?
if you want the "average rate of change" then you want to construct a line from \[(0, 2^2+1) = (0,5) \rightarrow (2,2^4+1) = (2,17)\] and figure out the slope
Average rate of change between x = 0 to x = 2 is: { f(2) - f(0) } / { 2 - 0 }
@polaris_s0i do i solve it ? and @ranga so the slope is 2?
Yeah @ranga has it \[slope = \frac{y_2 - y_1}{x_2-x_1}\] and slope is the average rate of change.
what are the two y
because there are 2 points: \[(x_1,y_1) = (0,5) and (x_2,y_2)=(2,17)\]
f(x) = 2^(x+2) + 1 f(2) = 2^(2+2) + 1 = 2^4 + 1 = 17 f(0) = 2^(0+2) + 1 = 2^2 + 1 = 5 Average rate of change between x = 0 to x = 2 is: { f(2) - f(0) } / { 2 - 0 } = (17-5) / 2 = 12 / 2 = 6
what is this process called @ranga
I don't know if there is any special name. What it does, is approximates a straight line between the two points instead of a curve and finds the slope of the straight line. That is the average rate of change. The derivative gives the instant rate of change at a point. Thus, f'(2) is the instantaneous rate of change at x = 2. f'(0) is the instantaneous rate of change at x = 0. But (f(2) - f(0)) / (2-0) is the average rate of change between x = 0 and x = 2. Graphically it is:
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you are calculating the distance in y direction over the distance in x. rise over run.
so slope equals 6
yes.
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