instantaneous rate of change?

Question

Find the instantaneous rate of change of y= X^2-4X at X=1. Use either numeric or limit method.

can you please explain in detail how you got your answer so that I can actually understand the stuff and do it by myself instead of just coping your thing.
just talk a little bit about instantaneous rate of change and limit method, mumeric method please...

Answer 1

limit method was a pain 50 years ago. I avoid it whenever I can.

dy/dx is the instantaneous rate of change

dy/dx = 2x-4 evaluate at x=1

dy/dx = -2

Answer 2

well I haven't used those methods in a while, but I believe they are basically like this.

When you have a secant line with points that approach the limit of what x is the secant lines slope will be approximately the same as the tangent one which would be the derivative.

so I'm going to pick a point that is .1 above and .1 below x=1 and then determine the slope of the line connecting these two points.

so for a start lets plug in .9 &1.1 into the function
(.9)²-4(.9) = -2.79
(1.1)²-4(1.1) =-3.19

ok you would have
-3.19--2.79/.2 which equals -2

Using more advance calculus will confirm that f'(1) is indeed -2.

Answer 3

Hello,

Let's look at the limit method.

1) We want the lim as t approaches zero of [f(x+t) - f(x)]/t .


so we have f(x+t) = (x+t)^2 - 4(x+t) = x^2 +2xt +t^2 -4x - 4t

and then -f(x) = -x^2 + 4x so using 1) we have

(2xt + t^2 -4t)/t = 2x + t -4

Now we want the limit of this as t approaches zero giving us

2x - 4

Putting in the point x = 1 gives the instantaneous rate of change = 2(1) - 4 = -2.

Hope This Helps!!

Answer 4

Crap, my calc sort in reality went over this at recent and that i have were given already forgotten (i imagine of). yet i'm fairly particular to discover the fee of substitute, you may take the second one by-product of the function, so acceptable right here it is going: f(x) = x / (x-4) f'(x) = [(x-4)(a million) - (x)(a million)] / (x-4)^2 = 4 / (x-4)^2 f''(x) = [(x-4)^2(0) - (4)(2(x-4) * (a million))] / (x-4)^4 = -8x + 32 / (x-4)^4 = -8(x - 4) / (x-4)^4 = -8 / (x-4)^3 desire that permits, in case you want me to describe the quotient rule and all, in reality tell me.

Answer 5

rate of change of y with respect to x is DEFINED as dy/dx.
Going back a number of years, when something is DEFINED, I accept------because someone more able than me was able to make the definition!
In this example:-
f(x) = y = x² - 4x
f ` (x) = dy/dx = 2x - 4
f ` (1) = 2 - 4 = - 4
-4 is the rate of change of y with respect to x.
If y is distance then dy/dx = velocity and d²y / dx² is acceleration.