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#.