Question

Let f (x) be a continuous function. If f (1) = 3 and f ′(x) ≤ −3

for all x ∈ R, what is the maximum possible value of f (5)? Briefly
explain your answer.

Answer 1

The maximum possible value of f(5) for the continuous function f(x) with f(1) = 3 and f′(x) ≤ -3 for all x ∈ R can be determined using the Mean Value Theorem. The maximum possible value of f(5) is 15.

Step-by-step explanation:

The maximum possible value of f(5) for the continuous function f(x) with f(1) = 3 and f′(x) ≤ -3 for all x ∈ R can be determined using the Mean Value Theorem. The Mean Value Theorem states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one point c in the open interval (a, b) such that the derivative of the function at c is equal to the average rate of change of the function over the interval [a, b]. In this case, we can consider the interval [1, 5] which satisfies the conditions of the Mean Value Theorem and find a value c such that f′(c) = (f(5) - f(1))/(5 - 1). Since f′(x) ≤ -3 for all x ∈ R, we can conclude that the maximum possible value of f(5) would be 3 - (-3) * 4 = 15.