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# f(x)=2x^3-7x^2+4x+1 Please Help?

I'm trying to find the Instantaneous rate of change using

f(a+h)-f(a)/(a+h)-a

I ended up with 10+2h+h²

I don't understand how to get the instantaneous rate of change from 10+2h+h² if i even did it right.

Using dh/dt=2x^3-7x^2+4x+1

=6x^2-14x+4

=6(1)^2-14(1)+4

=(-4)

i understand that h is the limit but how do i determine the limit if in not told?

sorry forgot to add x=1

### 1 Answer

- kbLv 71 decade agoFavorite Answer
Since f(x) = 2x^3 - 7x^2 + 4x + 1,

f(1) = 2 - 7 + 4 + 1 = 0

f(1 + h) = 2(1 + h)^3 - 7(1 + h)^2 + 4(1 + h) + 1

= 2(1 + 3h + 3h^2 + h^3) - 7(1 + 2h + h^2) + (4h + 4) + 1

= (2 + 6h + 6h^2 + 2h^3) + (-7 - 14h - 7h^2) + 4h + 5

= 2h^3 - h^2 - 4h.

So, we get

[f(1 + h) - f(1)] / h

= [( 2h^3 - h^2 - 4h) - 0] / h

= h * (2h^2 - h - 4) / h

= 2h^2 - h - 4.

To get the instantaneous rate of change, we let h = 0.

So, the answer is -4.

I hope that helps!