How do you find the instantaneous rate of change for $h(t)=-5t^2+20t+1$ for t=2?

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Answer 1 · 2018-04-23T03:29:29

$h'(t) = -10 t + 20$ so $h'(2)=-10(2) + 20 =0,$ so zero is the instantaneous rate of change at $t=2$ .

Answer 2 · 2018-04-23T03:52:43

No change.

The instantaneous rate of change for a function is the derivative of that function.

Given: $h(t)=-5t^2+20t+1$ .

The derivative is $h'(t)=-10t+20$ .

So at $t=2$ , the rate of change is:

$h'(2)=-10*2+20$

$=-20+20$

$=0$

Therefore, there is no change at $t=2$ .

How do you find the instantaneous rate of change for h(t)=-5t^2+20t+1h(t)=-5t^2+20t+1 for t=2?