The graph of f '(x) is continuous and decreasing with an x-intercept at x = 2. Which of the following statements must be true? (5 points) The graph of f is always concave down. The graph of f is always increasing. The graph of f has an inflection point at x = 2. The graph of f has a relative minimum at x = 2.

Answer:The graph of f is always concave down ⇒ the first answerStep-by-step explanation:* Lets explain how to solve the problem- Remember that : If  f(x)  is a function then the solutions to the  equation f′(x) = 0 gives the maximum and minimum values to f(x)- The value of  x  gives maximum if f′′(x) is negative and minimum if   f′′(x) is positive.- Inflection points of the function  f(x) are found the solutions of the  equation  f′′(x) = 0* Lets solve the problem- The graph of f'(x) is continuous means that the graph is unbroken line - The graph of f'(x) decreasing with an x-intercept at x = 2 means  f'(2) = 0- The differentiation of a function equal to zero at the critical point   (minimum or maximum) of the function∵ f'(x) = 0 at x = 2∴ The x-coordinate of the critical point of f(x) is 2- If the differentiation of the function is decreasing, then the critical  point of the function is maximum point∵ f'(x) is decreasing∴ The critical point of the f(x) is maximum point- That means the slope of curve is negative∴ The graph of f is concave down at x = 2* The right answer is the graph of f is always concave down