Tangent Lines and Rates of Change — Question 1

Question 1 - Solution

We are given: $$f(x) = \sqrt{x + 4}$$

Step 1: Slope of secant line from $x = 0$ to $x = 5$. $$f(0) = \sqrt{4} = 2, \quad f(5) = \sqrt{9} = 3$$ $$\text{Secant slope} = \frac{3 - 2}{5 - 0} = \frac{1}{5}$$

Step 2: Derivative of $f(x)$. $$f'(x) = \frac{1}{2\sqrt{x + 4}}$$ $$f'(2) = \frac{1}{2\sqrt{6}} \approx 0.204$$

Step 3: Point at $x = 2$. $$f(2) = \sqrt{6}$$

Step 4: Tangent line at $x = 2$: $$y - \sqrt{6} = \frac{1}{2\sqrt{6}}(x - 2)$$

Comparison: The slope of the tangent line at $x = 2$ is slightly greater than the secant slope over the interval, showing that the graph is concave down.