Question

the figure above shows the graph of the differentiable function (f) for (1leq xleq8) 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

1. 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:

1.

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 \([a,b]\). In the given problem, \(a = 1\), \(b = 8\). We are looking for the number of times the tangent line to the curve \(y = f(x)\) has the same slope as the secant line through \((1,f(1))\) and \((8,f(8))\). Geometrically, this is equivalent to the number of times the tangent line to the curve is parallel to the secant line. By observing the graph, we can count the number of points where the tangent line is parallel to the secant line. We see that there are three such points.

Step1: Find the velocity function

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

Step2: Find the values of \(t\) when \(v(t)=0\)

Set \(v(t)=4t^{3}-12t^{2}=4t^{2}(t - 3)=0\). Solving this equation gives \(t = 0\) or \(t=3\). 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 \(a(t)=v^{\prime}(t)=\frac{d}{dt}(4t^{3}-12t^{2})=12t^{2}-24t\).

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

2.