16. \( f(x)=\sqrt{x+8} \)

To solve a study problem involving the function \( f(x) = \sqrt{x + 8} \), you would typically follow these steps: 1. **Understanding the Function**: Recognize that \( f(x) \) is the square root of \( x + 8 \). This means that the function is only defined for non-negative values of \( x + 8 \) because the square root of a negative number is not a real number. 2. **Domain**: The domain of \( f(x) \) is all real numbers \( x \) such that \( x + 8 \geq 0 \). Solving this inequality gives us \( x \geq -8 \). So, the domain is \( [-8, \infty) \). 3. **Range**: Since the square root function always returns non-negative values, the range of \( f(x) \) is \( [0, \infty) \). 4. **Intercepts**: To find the x-intercept, set \( f(x) = 0 \) and solve for \( x \): \[ \sqrt{x + 8} = 0 \implies x + 8 = 0 \implies x = -8 \] So, the x-intercept is at \( x = -8 \). 5. **Critical Points**: To find critical points, we need to find where the derivative of \( f(x) \) is zero or undefined. The derivative of \( f(x) \) is: \[ f'(x) = \frac{1}{2\sqrt{x + 8}} \] The derivative is never undefined for \( x \geq -8 \), and it is zero only when \( x = -8 \), which is not in the domain of \( f(x) \) because the derivative is not defined at \( x = -8 \) (the point where the function is not differentiable). 6. **Increasing/Decreasing**: Since the derivative \( f'(x) \) is positive for all \( x > -8 \), the function is increasing on its entire domain. 7. **Concavity**: The second derivative of \( f(x) \) is: \[ f''(x) = -\frac{1}{4(x + 8)^{3/2}} \] Since \( f''(x) \) is negative for all \( x > -8 \), the function is concave down on its entire domain. 8. **Graphing**: You can graph \( f(x) \) to visualize these properties. The graph will be a curve starting at the point \( (-8, 0) \) and increasing and concave down as \( x \) increases. If you have a specific problem related to this function, please provide the details, and I can help you solve it.