Question

Please help please I dont know how to do this

Answer 1

`f(x)=2x^2-3x+2`

To find the instantaneous rate of change:

1. Find the derivate of f(x):

`\begin{gathered} (d)/(dx)ax^n=a*nx^(n-1) \\ \\ (d)/(dx)a=0 \\ \\ \\ (d)/(dx)f(x)=2*2x^(2-1)-3x^(1-1)+0 \\ \\ (d)/(dx)f(x)=4x-3 \\ \\ f^(\prime)(x)=4x-3 \end{gathered}`

2. Evaluate the derivate of f(x) for x=0:

`\begin{gathered} f^(\prime)(0)=4(0)-3 \\ f^(\prime)(0)=0-3 \\ f^(\prime)(0)=-3 \end{gathered}`

The instantaneus rate of change at x=0 of f(x) is the value of the derivate f'(x) evaluated in x=0.

Then, the instantaneus rate of change at x=0 is -3