Let \( f \) be the function given by \( f(x)=5 x^{\frac{2}{3}} \). What is the approximation for \( f(1.15) \) found by using the line tangent to the graph of \( f \) at \( x=1 \) ? Answer Attempt 1 out of 2 Tangent Line: \( y= \) Approximation: \( f(1.15) \approx \)

To find the tangent line to the graph of \( f(x) = 5x^{\frac{2}{3}} \) at \( x = 1 \), we first need to calculate the derivative of \( f \). 1. **Find \( f(1) \)**: \[ f(1) = 5(1)^{\frac{2}{3}} = 5 \] 2. **Find \( f'(x) \)** using the power rule: \[ f'(x) = 5 \cdot \frac{2}{3} x^{-\frac{1}{3}} = \frac{10}{3} x^{-\frac{1}{3}} \] 3. **Calculate \( f'(1) \)**: \[ f'(1) = \frac{10}{3}(1)^{-\frac{1}{3}} = \frac{10}{3} \] Now, we can write the equation of the tangent line \( y = f(1) + f'(1)(x - 1) \): \[ y = 5 + \frac{10}{3}(x - 1) \] To express this in slope-intercept form: \[ y = 5 + \frac{10}{3}x - \frac{10}{3} = \frac{10}{3}x + \frac{5}{3} \] Now, to approximate \( f(1.15) \): 1. **Substitute \( x = 1.15 \)** into the tangent line equation: \[ y \approx \frac{10}{3}(1.15) + \frac{5}{3} \] \[ y \approx \frac{10 \times 1.15}{3} + \frac{5}{3} = \frac{11.5}{3} + \frac{5}{3} = \frac{16.5}{3} \approx 5.5 \] Putting these all together: Tangent Line: \( y = \frac{10}{3}x + \frac{5}{3} \) Approximation: \( f(1.15) \approx 5.5 \)