Find the instantaneous rate of change for the function at the given value.
f(x)=5x+9 at x=2
the slope is 5 everywhere !
Ah, Now we have a parabola.
That is not pre- calc
It is calc
3t^2+6 at t=4???
if y = 3 t^2 + constant
then
dy/dt = slope = 6 t
at t = 4
dy/dt = 6*4 = 24
now if you are not supposed to know that you could plug in
at t = 4, y = 3*16 + 6 = 54
at t = 4.1 y = 3* 16.81 + 6 = 56.43
change in y = 2.43
change in x = 0.1
slope = 2.43 / .1 = 24.3
if you chose t = 4.001, you would get closer to 24
Since this is a linear function, the instantanious rate of change will be a constant,
namely the slope of the line, which is 5
I will be 5 no matter what x you choose
Thank you so to show work would I just plug in multiple different x-values
3t^2+6 at t=4???
d/dt (3t^2+6) = 6t
so, at t=4, that would be 24