$f(x) = 4+ 3x -x^{2}, \space \dfrac{f(3+h) – f(3)}{h}$

This question belongs to the calculus domain, and the aim is to understand the difference quotient and the practical application where it is being used.

The difference quotient is the term for the expression:

$\dfrac{f(x+h)-f(h)}{h}$

Where, when the limit h approaches $\rightarrow$ 0, delivers the derivative of the function $f$. As the expression itself explains that it is the quotient of the difference of the values of the function by the difference of the affiliated values of its argument. The rate of change of the function throughout length $h$ is called as the difference quotient. The limit of the difference quotient is the instantaneous rate of change.

In numerical differentiation the difference quotients are used as approximations, In time discretization, the difference quotient may also find relevance. Where the width of the time step is input as the value $h$.

Given the function $f(x)$ is:

$f(x) = 4+3x-x^{2}$

The difference quotient is given as:

$\dfrac{f(3+h) – f(3)}{h}$:

First, we will compute the expression for $f(3+h)$:

$f(x) = 4+3x-x^{2}$

$f(3+h) = 4+ 3(3+h)- (3+h)^{2}$

Expanding $(3+h)^{2}$ using the formula $(a+b)^2 = a^2 + b^2 + 2ab$

$f(3+h) = 4+ 9+3h- (3^2 + h^2 + 2(3)(h)$

$f(3+h) = 4+ 9+3h- (3^2 + h^2 + 2(3)(h))$

$f(3+h) = 13+3h – (9+ h^2 + 6(h))$

$f(3+h) = 13+3h -9 -h^2 -6(h))$

$f(3+h) = 4 -3h -h^2$

Now computing the expression for $f(3)$:

$f(x) = 4+3x- x^{2}$

$f(3) = 4+3(3)- (3)^{2}$

$f(3) = 4+9- 9$

$f(3) = 4$

Now insert the expressions in the difference quotient:

$= \dfrac{f(3+h) – f(3)} {h}$

$=\dfrac{(4 -3h -h^2) – 4} {h}$

$=\dfrac{4 -3h -h^2 -4} {h}$

$= \dfrac{h(-3 -h)} {h}$

$= -3 -h$

The difference quotient $\dfrac{f(3+h) – f(3)}{h}$ for the function $f(x) = 4+3x-x^{2}$ is $-3 -h$.

Example

Given the function:

$f(x) = -x^3, \space \dfrac{f(a+h) – f(a)}{h}$

Given the function $f(x)$ is:

$f(x) = -x^ {3}$

The difference quotient is given as:

$\dfrac{f(a+h) – f(a)} {h}$

Firstly we will compute the expression for $f(a+h)$:

$f(x) = -x^{3}$

$f(a+h) = – (a+h)^ {3}$

Expanding $(3+h)^{2}$ using the formula $(a+b)^3 = a^3 + b^3 + 3a^2b + 3ab^2$

$f(a+h) = – (a^3 + h^3 + 3a^2h + 3ah^2)$

Now computing the expression for $f(a)$:

$f(x) = – x^{3}$

$f(a) = -a^{3}$

Now insert the expressions in the difference quotient:

$= \dfrac{f(a+h) – f(a)}{h}$

$=\dfrac{- (a^3 + h^3 + 3a^2h + 3ah^2) – (-a^{3})} {h}$

$=\dfrac{ -a^3 -h^3 -3a^2h -3ah^2 +a^{3}} {h}$

$=\dfrac{ -h^3 -3a^2h -3ah^2 } {h}$

$=\dfrac{h( -h^2 -3a^2 -3ah) } {h}$

$= -3a^2 -3ah -h^2$

The difference quotient $\dfrac{f(a+h) – f(a)}{h}$ for the function $f(x) = -x^{3}$ is $-3a^2 -3ah -h^2$.