How to solve this problem.?

Question

Find the average rate of change of the function y=f(x)=x^2 between x=2 and x=5:find its instantaneous rate of change at x=2.

Answer 1

Average rate of change is a fancy term for slope.

(5^2 - 2^2)/(5-2) = 21/3 = 7

Instantaneous rate of change requires deriving a function

f(x) = x^2
f'(x) = 2x

Now plug in 2: 2*2 = 4

Answer 2

Find the average rate of change of the function y=f(x)=x^2 between x=2 and x=5:find its instantaneous rate of change at x=2.
Ans:
Average rate= {f(5) - f(2)}/(5-2)
Average rate= {5^2 - 2^2}/3
Average rate= {25 - 4}/3=21/3=7..................i
Instantaneous rate of change= {f(2+h) -f(2)}/{(2+h) -2} where h is very small
Instantaneous rate of change= {f(2+h) -f(2)}/h
={(2+h)^2-4}/h
={4+4h+h^2-4}/h
= {h(4+h)}/h
=4+h
Taking the limit when h tends to zero
Instantaneous rate of change= 4...............ii

Answer 3

Avg rate of change would be f(x) at 2 and 5 i.e.
5^2-2^2/(5-2)=25-4/3=21/3=7

Instantaneous rate of change=dy/dx
=2x
therefore at x=2=>dy/dx=2*2=4.

Answer 4

avg. rate of change =(net change between the limits)/(difference of x)=(5^2-2^2)/(5-2)=7

dy/dx=2x, so instantaneous rate of change at x=2 will equal to 4

Answer 5

y = x^2 is a parabola opening upward and at x=2, y=4 and at x=5,y=25 avg rate is (25-4)/3,instantaneous rate at x=2 is 4.We get inst.rate by differentiating the curve y=x^2 we get 2x substituting x=2 we get 4.

Answer 6

Avg = (5^2-2^2)/5-2
= 21/3
= 7

Instantaneous = dy/dx
= 2x
= 2*2
= 4

Answer 7

avg rate of change=(5^2+2^2)/7=29/7
instantaneous calcuclate dy/dx=2x=2(2)=4

Answer 8

avg rate f change: (y2-y1)/(x2-x1) = (5^2-2^2)/(5-3) = 5+3 = 8.

Inst rate f change = d(x^2)|
dx |(x=2)
= 2x = 2*2 = 4.

ANS:
avg rate f change=8
inst rate f change=4.

Answer 9

dy/dx for instentenuous rate of change then put x=2.
y2-y1/x2-x1 for average rate of change.

Answer 10

My maths is not so good........ so I cant help u out.......sorry........... Anyways Gud luck............