If f(t)=16t², obtain an expression for f(t+h)-f(t)/h?

Answer 1

f`(x)=[16(t+h)^2-16t^2]/h

=[16(t^2+2th+h^2)-16t^2]/h

=(16t^2+32th+16h^2-16t^2)/h

=(32th+16h^2)/h

=32t+16h as h goes to zero, 32t is your answer

Note, the power rule is even easier: 16*2 (exponent), then subrtact one from the exponent, leaving 1, or t, so it also = 32t

Answer 2

Did you mean [f(t+h) - f(t)]/h?

Then

[16(t+h)^2 - 16t^2]/h = (16t^2 + 2*16ht +16h^2 -16t^2)/h = (32ht +16 h^2)/h =
= 32t+16h


I assume that it is a calculus question and |h| is very small, but still not zero.

Answer 3

Puggy rarely makes a slip but he has here. Four lines from the end it should be
[16t^2 + 32th + 16h^2 - 16t^2]/h
= [32th + 16h^2]/h
=32t + 16h

Answer 4

f(t) = 16t²

[ f(t + h) - f(t) ] / h =

[ 16(t + h)² - 16t² ] / h =

[ 16(t² + 2th + h²) - 16t² ] / h =

[ 16t² + 32th -16h² ] / h =

32th / h =

Canceling the h, we get

32t