The problem is asking for the smallest possible value of f(4) given that f(x) is continuous and differentiable, f(1) = 10, and f’(x) ≥ 3 for 1 ≤ x ≤ 4.

The function f(x) is differentiable and continuous, and its derivative f’(x) is greater than or equal to 3 in the interval [1, 4]. This means that the function f(x) is increasing at a rate of at least 3 units per unit increase in x.

Since f(1) = 10, and the function increases at a rate of at least 3 units per unit increase in x, the smallest possible value of f(4) would be achieved if the function increased at exactly this minimum rate over the interval from x = 1 to x = 4.

The increase in x over this interval is 4 - 1 = 3 units. If the function increases at a rate of 3 units per unit increase in x, then the increase in f(x) over this interval would be 3 * 3 = 9 units.

Therefore, the smallest possible value of f(4) would be f(1) + 9 = 10 + 9 = 19.

So, the answer is D. 19.

Question

The problem is asking for the smallest possible value of f(4) given that f(x) is continuous and differentiable, f(1) = 10, and f’(x) ≥ 3 for 1 ≤ x ≤ 4.

The function f(x) is differentiable and continuous, and its derivative f’(x) is greater than or equal to 3 in the interval [1, 4]. This means that the function f(x) is increasing at a rate of at least 3 units per unit increase in x.

Since f(1) = 10, and the function increases at a rate of at least 3 units per unit increase in x, the smallest possible value of f(4) would be achieved if the function increased at exactly this minimum rate over the interval from x = 1 to x = 4.

The increase in x over this interval is 4 - 1 = 3 units. If the function increases at a rate of 3 units per unit increase in x, then the increase in f(x) over this interval would be 3 * 3 = 9 units.

Therefore, the smallest possible value of f(4) would be f(1) + 9 = 10 + 9 = 19.

So, the answer is D. 19.

Knowee AI · Accepted Answer

The problem is asking for the smallest possible value of f(4) given that f(x) is continuous and differentiable, f(1) = 10, and f’(x) ≥ 3 for 1 ≤ x ≤ 4.

The function f(x) is differentiable and continuous, and its derivative f’(x) is greater than or equal to 3 in the interval [1, 4]. This means that the function f(x) is increasing at a rate of at least 3 units per unit increase in x.

Since f(1) = 10, and the function increases at a rate of at least 3 units per unit increase in x, the smallest possible value of f(4) would be achieved if the function increased at exactly this minimum rate over the interval from x = 1 to x = 4.

The increase in x over this interval is 4 - 1 = 3 units. If the function increases at a rate of 3 units per unit increase in x, then the increase in f(x) over this interval would be 3 * 3 = 9 units.

Therefore, the smallest possible value of f(4) would be f(1) + 9 = 10 + 9 = 19.

So, the answer is D. 19.

If f(x) is continuous & differentiable,f(1) = 10 and f’(x) ≥ 3 in 1 ≤ x ≤ 4 thenthe smallest value of f(4) can beA. 10 B. 13C. 14 D. 19

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