Question

Calc, I need help with the first one so that I know how to do the rest of them

https://www.dropbox.com/s/51u3qcz8i80tz98/Screenshot%202016-01-10%2013.19.45.png?dl=0

Answer 1

11. so my opinion, i think this way :

y=2x^2 you know x_0 = 0 and x_1 = 1

for x_0 = 0 result y= 0
for x_1 = 1 ---- y= 1

a) the average rate of change in this x_0,x_1 interval i think will be 1 unit

Answer 2

https://www.dropbox.com/s/0k9apf7nlhx9l5x/Screenshot%202016-01-10%2013.50.41.png?dl=0

Answer 3

*(for x_1 = 1 ---- y= 1) this should be 2*1 = 2.
He solved it right, just slight mistake.

Answer 4

problem is, I don't understand wha't the intuition behind this

Answer 5

Anyway what are you struggling with?

Answer 6

the simple average rate of change over an interval is is \(f'_{ave} = \dfrac{f_1 - f_0}{x_1 - x_0}\)

if you average the instantaneous rate over the same period you get \(f'_{ave} = \dfrac{1}{x_1 - x_0} \int_{x_0}^{x_1} f'(x) \; dx = \dfrac{f_1 - f_0}{x_1 - x_0}\) , ie the same.....

Answer 7

what is the equation of the SECANT line that is giving me this 'slope' ?

Answer 8

how would you find the equation of this secant line

Answer 9

\[y = m x + c\]
m is slope(average rate)
and c is the y intercept as you know.
so you will get \[\frac{ y-y0 }{ x-x0 } = \frac{ y1-y2 }{ x2-x1 }\]

Answer 10

i meant y2 - y1

Answer 11

so the derivative of the avg rate of change gives me the instantanious rate of change ?

Answer 12

no the avg rate of change over a very small region when x1 tends to x0 will give instantaneous rate of change

or simply when delta x is very small

Answer 13

y2-y1/x2-x1
is an approximation for f'

and approx. will be exactly f' when it is very small(limit).