1. **Problem statement:** Find the equation of the tangent line to the function $$f(x) = \sqrt{25 - x^2}$$ at the point where $$x = 4$$. 2. **Formula and rules:** The slope of the tangent line at a point is given by the derivative $$f'(x)$$ evaluated at that point. The tangent line equation in slope-intercept form is $$y = mx + b$$, where $$m = f'(x_0)$$ and $$b$$ is found using the point on the curve. 3. **Find the derivative:** $$f(x) = (25 - x^2)^{1/2}$$ Using the chain rule: $$f'(x) = \frac{1}{2}(25 - x^2)^{-1/2} \cdot (-2x) = \frac{-x}{\sqrt{25 - x^2}}$$ 4. **Evaluate the derivative at $$x=4$$:** $$f'(4) = \frac{-4}{\sqrt{25 - 16}} = \frac{-4}{\sqrt{9}} = \frac{-4}{3}$$ 5. **Find the point on the curve at $$x=4$$:** $$f(4) = \sqrt{25 - 16} = \sqrt{9} = 3$$ So the point is $$(4, 3)$$. 6. **Write the tangent line equation:** Using point-slope form: $$y - 3 = -\frac{4}{3}(x - 4)$$ Simplify: $$y - 3 = -\frac{4}{3}x + \frac{16}{3}$$ $$y = -\frac{4}{3}x + \frac{16}{3} + 3$$ Convert 3 to thirds: $$3 = \frac{9}{3}$$ So: $$y = -\frac{4}{3}x + \frac{25}{3}$$ --- 7. **Problem statement:** Find the equation of the tangent line to the function $$f(x) = x^2 + \sqrt{x}$$ at the point where $$x = 1$$. 8. **Find the derivative:** $$f(x) = x^2 + x^{1/2}$$ $$f'(x) = 2x + \frac{1}{2}x^{-1/2} = 2x + \frac{1}{2\sqrt{x}}$$ 9. **Evaluate the derivative at $$x=1$$:** $$f'(1) = 2(1) + \frac{1}{2\sqrt{1}} = 2 + \frac{1}{2} = \frac{5}{2}$$ 10. **Find the point on the curve at $$x=1$$:** $$f(1) = 1^2 + \sqrt{1} = 1 + 1 = 2$$ So the point is $$(1, 2)$$. 11. **Write the tangent line equation:** Using point-slope form: $$y - 2 = \frac{5}{2}(x - 1)$$ Simplify: $$y - 2 = \frac{5}{2}x - \frac{5}{2}$$ $$y = \frac{5}{2}x - \frac{5}{2} + 2$$ Convert 2 to halves: $$2 = \frac{4}{2}$$ So: $$y = \frac{5}{2}x - \frac{1}{2}$$