Question

1. the figure above shows the graph of the differentiable function f for 1 ≤ x ≤ 8 and the secant line through the points (1, f(1)) and (8, f(8)). for how many values of x in the closed interval 1, 8 does the instantaneous rate of change of f at x equal the average rate of change of f over that interval? (a) zero (b) three (c) four (d) five 2. for time t > 0, the position of a particle moving along the x - axis is given by x(t)=t^4 - 4t^3. what is the acceleration of the particle when the velocity of the particle is 0? (a) - 16 (b) 0 (c) 36 (d) 96

Explanation:

Step1: Recall the Mean - Value Theorem

The Mean - Value Theorem states that if \(y = f(x)\) is continuous on the closed interval \([a,b]\) and differentiable on the open interval \((a,b)\), then there exists at least one \(c\in(a,b)\) such that \(f^{\prime}(c)=\frac{f(b)-f(a)}{b - a}\), where \(f^{\prime}(c)\) is the instantaneous rate of change of \(f\) at \(x = c\) and \(\frac{f(b)-f(a)}{b - a}\) is the average rate of change of \(f\) over \([a,b]\). Here, \(a = 1\), \(b = 8\), and \(y=f(x)\) is differentiable on \([1,8]\). Geometrically, \(f^{\prime}(c)\) is the slope of the tangent line at \(x = c\) and \(\frac{f(8)-f(1)}{8 - 1}\) is the slope of the secant line through \((1,f(1))\) and \((8,f(8))\). We need to find the number of times the tangent line to \(y = f(x)\) has the same slope as the secant line.

Step2: Analyze the graph

By looking at the graph of \(y = f(x)\) on the interval \([1,8]\), we count the number of points where the tangent line to the curve \(y = f(x)\) is parallel to the secant line through the points \((1,f(1))\) and \((8,f(8))\). We can see that there are three such points.

for Question 2:

Step1: Find the velocity function

The velocity function \(v(t)\) is the derivative of the position function \(x(t)\). Given \(x(t)=t^{4}-4t^{3}\), using the power - rule \((x^{n})^\prime=nx^{n - 1}\), we have \(v(t)=x^{\prime}(t)=4t^{3}-12t^{2}\).

Step2: Find the times when velocity is 0

Set \(v(t)=0\), so \(4t^{3}-12t^{2}=4t^{2}(t - 3)=0\). Solving for \(t\), we get \(t = 0\) and \(t=3\). Since \(t>0\), we consider \(t = 3\).

Step3: Find the acceleration function

The acceleration function \(a(t)\) is the derivative of the velocity function. So \(a(t)=v^{\prime}(t)=(4t^{3}-12t^{2})^\prime=12t^{2}-24t\).

Step4: Evaluate the acceleration at \(t = 3\)

Substitute \(t = 3\) into the acceleration function: \(a(3)=12\times3^{2}-24\times3=12\times9-72 = 108 - 72=36\).

Answer:

B. Three