Use the limit definition of the derivative to find the slope of the tangent line to the curve \( f(x)=5 x^{2} \) at \( x=3 \).

To find the slope of the tangent line to the curve \( f(x) = 5x^2 \) at \( x = 3 \) using the limit definition of the derivative, follow these steps: ### 1. **Recall the Limit Definition of the Derivative** The derivative of \( f(x) \) at a point \( x = a \) is given by: \[ f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h} \] ### 2. **Apply the Definition to \( f(x) = 5x^2 \) at \( x = 3 \)** Substitute \( a = 3 \) into the formula: \[ f'(3) = \lim_{h \to 0} \frac{f(3 + h) - f(3)}{h} \] ### 3. **Compute \( f(3 + h) \) and \( f(3) \)** \[ f(3 + h) = 5(3 + h)^2 = 5(9 + 6h + h^2) = 45 + 30h + 5h^2 \] \[ f(3) = 5(3)^2 = 45 \] ### 4. **Subtract \( f(3) \) from \( f(3 + h) \)** \[ f(3 + h) - f(3) = (45 + 30h + 5h^2) - 45 = 30h + 5h^2 \] ### 5. **Set Up the Difference Quotient** \[ \frac{f(3 + h) - f(3)}{h} = \frac{30h + 5h^2}{h} = 30 + 5h \] ### 6. **Take the Limit as \( h \) Approaches 0** \[ f'(3) = \lim_{h \to 0} (30 + 5h) = 30 + 0 = 30 \] ### **Conclusion** The slope of the tangent line to the curve \( f(x) = 5x^2 \) at \( x = 3 \) is **30**. ### **Graphical Interpretation** This means that at the point \( (3, 45) \) on the curve, the tangent line has a slope of 30. If you were to draw this tangent line, it would rise 30 units vertically for every 1 unit it moves horizontally at the point of tangency.