Find the average rate of change of the function over the given interval. f(t) = 2t^2 − 1, [5, 5.1]

Compare this average rate of change with the instantaneous rates of change at the endpoints of the interval.

Average rate of change = 20.2

Rate of change at the left endpoint :   f' (5) = 4t = 20

Rate of change at the right  endpoint : f' (5.1) = 4*5.1 = 20.4

Step-by-step explanation:

The average rate of change of the function

F(t) = 2t^2 - 1 ,  [ 5,5.1 ]

solution

=   = [ 2 ( 26.01 - 25 ) / 0.1 ]

= 2.02 / 0.1 = 20.2

Rate of change at the left endpoint :   f' (5) = 4t = 20

Rate of change at the right  endpoint : f' (5.1) = 4*5.1 = 20.4

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}