question 7 (1 point)\nthe mean value theorem states that if a function f(x) is continuous on a,b and…

question 7 (1 point)\nthe mean value theorem states that if a function f(x) is continuous on a,b and differentiable on (a,b), then there exists a point c in (a,b) such that the slope of the tangent line at c equals:\nzero\nthe slope of the secant line between a and b\nthe average rate of change of f between 0 and c\nf(c)\nview hint for question 7\nquestion 8 (1 point)\nsuppose f(x) is differentiable on (-∞, ∞), f(-3) = 4, and f(x) ≤ 3 for all values of x. using mean value theorem, the largest f(4) can possibly be is:\n. (enter a number for you answer.)\nview hint for question 8

question 7 (1 point)\nthe mean value theorem states that if a function f(x) is continuous on a,b and differentiable on (a,b), then there exists a point c in (a,b) such that the slope of the tangent line at c equals:\nzero\nthe slope of the secant line between a and b\nthe average rate of change of f between 0 and c\nf(c)\nview hint for question 7\nquestion 8 (1 point)\nsuppose f(x) is differentiable on (-∞, ∞), f(-3) = 4, and f(x) ≤ 3 for all values of x. using mean value theorem, the largest f(4) can possibly be is:\n. (enter a number for you answer.)\nview hint for question 8

Answer

Explanation:

Step1: Apply the Mean Value Theorem formula

The Mean Value Theorem states that (f^{\prime}(c)=\frac{f(b)-f(a)}{b - a}). Here, (a=-3), (b = 4), so (f^{\prime}(c)=\frac{f(4)-f(-3)}{4-(-3)}=\frac{f(4)-4}{7}).

Step2: Use the condition on the derivative

We know that (f^{\prime}(x)\leq3) for all (x). Since (c) is in ((-3,4)) (by the Mean Value Theorem), (f^{\prime}(c)\leq3). Substituting (f^{\prime}(c)=\frac{f(4)-4}{7}) into the inequality (\frac{f(4)-4}{7}\leq3).

Step3: Solve the inequality for (f(4))

Multiply both sides of the inequality (\frac{f(4)-4}{7}\leq3) by (7): (f(4)-4\leq21). Then add (4) to both sides: (f(4)\leq21 + 4).

Answer:

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