Question

EXAMPLE 4 Find the derivative of the function
f(x) = x2 − 2x + 2
at the number a.

SOLUTION From the definition we have
f '(a) =
lim
h → 0
f(a + h) − f(a)
h
=
lim
h → 0

− 2(a + h) + 2

− [a2 − 2a + 2]
h
=
lim
h → 0
a2 + − 2a − 2h + 2 − a2 + 2a − 2
h
=
lim
h → 0

h
=
lim
h → 0



= .

Answer 1

To find the derivative of the function \( f(x) = x^2 - 2x + 2 \) at the number \( a \), we can follow the definition of the derivative.

The definition states:

\[
f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}
\]

1. First, we calculate \( f(a + h) \):

\[
f(a + h) = (a + h)^2 - 2(a + h) + 2
\]
Expanding this, we get:
\[
f(a + h) = a^2 + 2ah + h^2 - 2a - 2h + 2
\]
Combining like terms, we have:
\[
f(a + h) = a^2 - 2a + 2 + 2ah + h^2 - 2h
\]

2. Then we need to calculate \( f(a) \):

\[
f(a) = a^2 - 2a + 2
\]

3. Now we substitute \( f(a + h) \) and \( f(a) \) into the limit:

\[
f'(a) = \lim_{h \to 0} \frac{(a^2 - 2a + 2 + 2ah + h^2 - 2h) - (a^2 - 2a + 2)}{h}
\]

4. Simplifying the expression inside the limit:

\[
f'(a) = \lim_{h \to 0} \frac{(2ah + h^2 - 2h)}{h}
\]

5. Factor \( h \) out from the numerator:

\[
f'(a) = \lim_{h \to 0} \frac{h(2a + h - 2)}{h}
\]

6. Cancel \( h \) (assuming \( h \neq 0 \)):

\[
f'(a) = \lim_{h \to 0} (2a + h - 2)
\]

7. As \( h \) approaches \( 0 \):

\[
f'(a) = 2a - 2
\]

Thus, the derivative of the function \( f(x) = x^2 - 2x + 2 \) at the number \( a \) is:

\[
\boxed{2a - 2}
\]