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.

The graph of f is always concave down ⇒ the first answer

Step-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

d is the correct answer

step-by-step explanation:

x= 8 because 3x = 3 * 8, and 3*8 = 24

eve and marks ratios are equivalent

step-by-step explanation: