Problem 18 - Calculus Early Transcendentals

Question

a. Determine whether the Mean Value Theorem applies to the following functions on the given interval \([a, b]\). b. If so, find the point(s) that are guaranteed to exist by the Mean Value Theorem. c. For those cases in which the Mean Value Theorem applies, make a sketch of the function and the line that passes through \((a, f(a))\) and\((b, f(b)) .\) Mark the points \(P\) at which the slope of the function equals the slope of the secant line. Then sketch the tangent line at \(P\). $$f(x)=3 \sin 2 x ;[0, \pi / 4]$$

Step-by-Step Solution

Answer

If so, find the point c in the interval and sketch the function with its tangent and secant lines. Answer: Yes, the Mean Value Theorem applies to the function \(f(x) = 3 \sin(2x)\) on the interval \([0, \pi/4]\). The point c that satisfies the MVT is given by: $$c = \frac{1}{2} \arccos\left(\frac{2}{\pi}\right)$$ Sketch the function, the secant line that passes through \((0, 0)\) and \((\pi/4, 3)\), and the tangent line at point c.

Step 1: Check conditions for MVT

We need to check whether the function meets the necessary criteria for MVT to apply. The Mean Value Theorem states that if a function \(f(x)\) is continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then we can find a point \(c \in (a, b)\) such that \(f'(c) = \frac{f(b) - f(a)}{b - a}\). For the given function: $$f(x) = 3 \sin(2x),$$ it is continuous and differentiable for all real values of x since sine and cosine functions are continuous and differentiable everywhere. Thus, on the interval \([0, \pi/4]\), the function is continuous and differentiable.

Step 2: Find derivative of f(x)

Now that we know the condition for the MVT is satisfied, let's find the derivative of the function: $$f'(x) = \frac{d}{dx}(3 \sin(2x)) = 3 \cdot \frac{d}{dx}(\sin(2x))$$ Using the chain rule: $$f'(x) = 3 \cdot (2\cos(2x))$$ So, the derivative is: $$f'(x) = 6\cos(2x)$$

Step 3: Find the point that satisfies MVT

Now, we need to find a point \(c\) in the interval \((0, \pi/4)\) such that the condition of MVT is satisfied: $$f'(c) = \frac{f(\pi/4) - f(0)}{(\pi/4) - 0}$$ Calculating \(f(\pi/4)\) and \(f(0)\): $$f(\pi/4) = 3\sin(2(\pi/4)) = 3\sin(\pi/2) = 3(1) = 3$$ $$f(0) = 3\sin(2\cdot0) = 3\sin(0) = 0$$ Now we have: $$f'(c) = \frac{3 - 0}{\pi/4 - 0}$$ $$f'(c) = \frac{3}{\pi/4}$$ To find the value of \(c\): $$6\cos(2c) = \frac{12}{\pi}$$ $$\cos(2c) = \frac{2}{\pi}$$ $$2c = \arccos\left(\frac{2}{\pi}\right)$$ $$c = \frac{1}{2} \arccos\left(\frac{2}{\pi}\right)$$

Step 4: Sketch the function and lines

Now that we have the point \(c\) that satisfies MVT, we can sketch the function, the secant line that passes through \((a, f(a))\) and \((b, f(b))\), and the tangent line at point \(P\) where the slope of the function equals the slope of the secant line. Notice that for this problem, we do not need to find the actual point \(P = (c, f(c))\) or the equations of the tangent and secant lines, but only their sketches are required. So, sketch the function \(f(x) = 3\sin(2x)\) on the interval \([0, \pi/4]\), draw a secant line that passes through \((0, 0)\) and \((\pi/4, 3)\), and then draw the tangent line at the point \((c, f(c))\). Mark the point \(P\) at which the slope of the function equals the slope of the secant line.

Key Concepts

Continuous Function

A continuous function is a type of function that can be drawn without lifting your pen off the paper. This means there are no breaks, gaps, or jumps in the graph of the function. Continuity is a crucial condition for the Mean Value Theorem (MVT) to apply.

Now, considering the function given in the exercise, \(f(x) = 3\sin(2x)\), sine functions are known for being continuous everywhere. They do not have any gaps or interruptions throughout their domain. Therefore, in the interval \([0, \pi/4]\), our function is continuous. This indicates that it perfectly meets the first requirement of the MVT.

Without continuity on the closed interval, the theorem would not apply because you could have points on the graph where the function suddenly jumps from one value to another, making it impossible to find a point within that interval that equates the derivative to the slope of the secant line.

Differentiable Function

Differentiability is another crucial requirement for applying the Mean Value Theorem. A function is differentiable if it has a derivative at every point in its domain. In simpler terms, it means the function has a clear, calculable slope at all points, making it smooth and without any sharp corners or cusps.

For the function \(f(x) = 3\sin(2x)\), differentiability on the interval \((0, \pi/4)\) ensures that the function behaves nicely without sudden changes in direction. Since trigonometric functions like sine and cosine are both continuous and differentiable across all real numbers, \(f(x)\) is differentiable on this open interval.

It's important to remember that differentiability implies continuity, so if a function is differentiable, it's also continuous. But not all continuous functions are differentiable. This distinction is critical when analyzing why a function satisfies the conditions necessary for MVT to apply.

Derivative Calculation

Calculating the derivative is key to using the Mean Value Theorem effectively. The derivative tells us the slope of the tangent line at any given point on the function. For \(f(x) = 3\sin(2x)\), the derivative calculation involves a bit of the chain rule since we have two functions to consider: the constant and the trigonometric function within.

We differentiate using the chain rule to get:

\[ f'(x) = 3 \cdot 2 \cos(2x) = 6\cos(2x) \]

This derivative indicates how sharply the curve is changing at each point in our interval. In the context of MVT, this derivative will be used to find a specific point \(c\) in the interval where the instantaneous rate of change (the slope at that point) equals the average rate of change over the interval. The derivative thus plays a critical role in locating the point \(c\) where the tangent line is parallel to the secant line, fulfilling the MVT condition.

Secant and Tangent Lines

Secant and tangent lines are important concepts in calculus for understanding the geometry of functions. A secant line connects two points on a curve, providing the average rate of change between those points—essentially, how much the function has changed between those two points.

In the case of \(f(x) = 3\sin(2x)\) over the interval \([0, \pi/4]\), the secant line connects the endpoints \((0, f(0))\) and \((\pi/4, f(\pi/4))\). Its slope is given by \(\frac{f(\pi/4) - f(0)}{(\pi/4) - 0}\), which is the rise over run formula for the line.

Tangent lines, on the other hand, only touch the function at one point and have the same slope as the function at that point. For MVT, we want the tangent line at the point \(P = (c, f(c))\) to have the same slope as the secant line. This means that the instantaneous rate of change at \(c\) will equal the average rate of change across the interval, reflecting the core intuition of the Mean Value Theorem.