We are tasked with finding the difference quotient for the given function $f(x) = x^2 - x + 1$, specifically for:

$\frac{f(2+h) - f(2)}{h}, \quad h \neq 0.$

Step 1: Calculate $f(2+h)$

Substitute $x = 2+h$ into $f(x) = x^2 - x + 1$:

$f(2+h) = (2+h)^2 - (2+h) + 1.$

Simplify $(2+h)^2$: $(2+h)^2 = 4 + 4h + h^2.$

Now substitute back: $f(2+h) = 4 + 4h + h^2 - (2+h) + 1.$

Simplify further: $f(2+h) = 4 + 4h + h^2 - 2 - h + 1 = h^2 + 3h + 3.$

Step 2: Calculate $f(2)$

Substitute $x = 2$ into $f(x) = x^2 - x + 1$: $f(2) = 2^2 - 2 + 1 = 4 - 2 + 1 = 3.$

Step 3: Difference Quotient

Substitute $f(2+h)$ and $f(2)$ into the difference quotient formula: $\frac{f(2+h) - f(2)}{h} = \frac{(h^2 + 3h + 3) - 3}{h}.$

Simplify the numerator: $(h^2 + 3h + 3) - 3 = h^2 + 3h.$

So the difference quotient becomes: $\frac{f(2+h) - f(2)}{h} = \frac{h^2 + 3h}{h}.$

Step 4: Simplify

Factor $h$ out of the numerator: $\frac{h^2 + 3h}{h} = \frac{h(h+3)}{h}.$

Since $h \neq 0$, cancel $h$ from the numerator and denominator: $\frac{h(h+3)}{h} = h+3.$

Final Answer:

$\frac{f(2+h) - f(2)}{h} = h + 3, \quad h \neq 0.$

Let me know if you have any questions or want a deeper explanation!

Related Questions:

1. What is the difference quotient and why is it useful in calculus?
2. How would the result change if $f(x)$ was a cubic function instead?
3. How do you generalize the process of finding $\frac{f(a+h)-f(a)}{h}$?
4. Can the difference quotient be used to approximate derivatives?
5. How do you interpret the limit of the difference quotient as $h \to 0$?

Tip: Always simplify the numerator completely before canceling terms in the difference quotient!