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Computing f'(3) using the limit definition

Let f(x)=x2−6x+9f(x) = x^2 - 6x + 9.

Use the limit definition of the derivative, f′(a)=lim⁡h→0f(a+h)−f(a)h,f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}, to compute f′(3)f'(3) for f(x)=x2−6x+9f(x) = x^2 - 6x + 9. You must show the limit computation explicitly — do not use shortcut differentiation rules.

Hint

First find f(3)f(3) and f(3+h)f(3+h), substitute into the difference quotient, expand the numerator, simplify by canceling a factor of hh, then take the limit as h→0h \to 0.

Step-by-step solution

First compute f(3)=32−6(3)+9=0f(3) = 3^2 - 6(3) + 9 = 0. Next compute f(3+h)=(3+h)2−6(3+h)+9f(3+h) = (3+h)^2 - 6(3+h) + 9. Expanding gives (9+6h+h2)−18−6h+9=h2(9 + 6h + h^2) - 18 - 6h + 9 = h^2. So the difference quotient is f(3+h)−f(3)h=h2−0h=h\frac{f(3+h) - f(3)}{h} = \frac{h^2 - 0}{h} = h, valid for h≠0h \neq 0.

Now take the limit as h→0h \to 0: f′(3)=lim⁡h→0h=0.f'(3) = \lim_{h \to 0} h = 0. Therefore the instantaneous rate of change of ff at x=3x = 3 is 00, which also matches the fact that f(x)=(x−3)2f(x) = (x-3)^2 has a minimum (horizontal tangent) at x=3x = 3.

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