Tangent Line 0C7C8F

1. **State the problem:** Find the equation of the tangent line to the function $f(x) = x^2$ at the point where $x = 3$ using the limit definition of the derivative. 2. **Find the point on the curve:** Calculate $f(3)$: $$f(3) = 3^2 = 9$$ So the point is $(3, 9)$. 3. **Recall the limit definition of the derivative:** $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ 4. **Apply the limit definition to $f(x) = x^2$:** $$f'(x) = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h}$$ Expand the numerator: $$(x+h)^2 - x^2 = x^2 + 2xh + h^2 - x^2 = 2xh + h^2$$ So, $$f'(x) = \lim_{h \to 0} \frac{2xh + h^2}{h}$$ Simplify by canceling $h$: $$f'(x) = \lim_{h \to 0} \frac{\cancel{h}(2x + h)}{\cancel{h}} = \lim_{h \to 0} (2x + h)$$ 5. **Evaluate the limit:** $$f'(x) = 2x$$ 6. **Find the slope at $x=3$:** $$f'(3) = 2 \times 3 = 6$$ 7. **Write the equation of the tangent line using point-slope form:** Point: $(3, 9)$, slope: $6$ $$y - 9 = 6(x - 3)$$ 8. **Simplify the equation:** $$y - 9 = 6x - 18$$ $$y = 6x - 9$$ **Final answer:** The equation of the tangent line to $f(x) = x^2$ at $x=3$ is $$y = 6x - 9$$