Question

Consider the function: F(x) = x^2 - 4x +1a) Simplify the expression : f(x+h) - f(x) / h , for the given function.b) What will happen to the expression in part a when h becomes very close to zero?c) Use the simplified expression from part a to estimate the instantaneous rate of change at: x=1 , x=2, and x=3. Explain what these results tell you about the function.

Answer 1

$\begin{gathered} a)2x+h-4 \\ b)\lim _{h\to0}2x-4 \\ c)x=-1,f^{\prime}(1)=-2 \\ x=2,f^{\prime}(2)=0 \\ x=3,f^{\prime}(3)=2 \end{gathered}$

1) Considering that quadratic equation, we can find and simplify this expression that presents the limit of this function as h approaches 0.

a)

$\begin{gathered} \frac{f(x+h)-f(x)}{h}= \\ \frac{(x+h)^2-4(x+h)+1-(x^2-4x+1)}{h} \\ \frac{x^2+2xh+h^2-4x+4h+1-x^2+4x-1}{h} \\ \frac{2xh+h^2-4h}{h} \\ \frac{h\left(2x+h-4\right)}{h} \\ 2x+h-4 \end{gathered}$

b) When h becomes very close to zero, we can find this:

$\begin{gathered} 2x+0-4 \\ 2x-4 \end{gathered}$

In other words, we find the limit of this function (or the first derivative) as h approaches 0.

c) Let's use this simplified expression to find the instantaneous rate of change at x=1,x=2, and x=3

$\begin{gathered} f^{\prime}(x)=\lim _{h\to0}2x-4 \\ f^{\prime}(1)=\lim _{h\to0}2x-4=2(1)-4=-2 \\ f^{\prime}(2)=\lim _{h\to0}2x-4=2(2)-4=0 \\ f^{\prime}(3)=\lim _{h\to0}2x-4=2(3)-4=2 \end{gathered}$