Question

question 7 (1 point)
the 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:
zero
the slope of the secant line between a and b
the average rate of change of f between 0 and c
f(c)
view hint for question 7
question 8 (1 point)
suppose 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:
. (enter a number for you answer.)
view hint for question 8

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:

\(25\)