Problem 23 - The Calculus with Analytic Geometry

Question

(a) Draw a sketch of the graph of the given function on the indicated interval; (b) test the three conditions (i), (ii), and (iii) of the hypothesis of Rolle's theorem and determine which conditions are satisfied and which, if any, are not satisfied; and (c) if the three conditions in part (b) are satisfied, determine a point at which there is a horizontal tangent line. $$ f(x)=\left|9-4 x^{2}\right| ;\left[-\frac{3}{2}, \frac{3}{2}\right] $$

Step-by-Step Solution

Answer

Rolle's theorem does not apply because the function is not differentiable at all points in the interval.

Step 1 - Understand the function and interval

The given function is \(f(x) = |9 - 4x^2 |\). The interval to consider is \(\bigg[-\frac{3}{2}, \frac{3}{2}\bigg] \).

Step 2 - Draw the sketch of the function

Plot the function \( f(x) = |9 - 4x^2| \) on the interval \( \bigg[-\frac{3}{2}, \frac{3}{2} \bigg] \). The function is a downward parabola flipped at \(|9 - 4x^2| = 0 \). Draw this graph carefully, noting that it will reach a minimum of 0 where \(x = \pm\frac{3}{2} \).

Step 3 - Check continuity of the function

Analyze if the function \(f(x)\) is continuous on the closed interval \( \bigg[-\frac{3}{2}, \frac{3}{2} \bigg] \). \(f(x) = |9 - 4x^2|\) is composed of polynomials, which are continuous, hence \(f(x)\) is continuous on the given interval.

Step 4 - Check differentiability

Check if \(f(x)\) is differentiable on the open interval \( \bigg(-\frac{3}{2}, \frac{3}{2} \bigg] \). The absolute value function can have a cusp which affects differentiability. The slope changes sign at \(x=0\).

Step 5 - Check if endpoints give the same value

Evaluate the function at the endpoints: \( f(-\frac{3}{2}) = f(\frac{3}{2}) = 0\). The function values at the endpoints are equal.

Step 6 - Determine whether Rolle's theorem applies

Although the function is continuous and it has equal values at the endpoints of the interval, it is not differentiable at \(x=0\) because of the cusp. Therefore, Rolle's theorem does not apply.

Key Concepts

Continuity

To understand Rolle's Theorem, we need to start with the concept of continuity. In simple terms, a function is continuous if you can draw it without lifting your pencil from the paper. For the function given in our exercise, \(f(x) = |9 - 4x^2|\), we need to check if it is continuous on the interval \(\bigg[-\frac{3}{2}, \frac{3}{2}\bigg]\).

Since the function is composed of polynomials inside the absolute value, and we know that polynomials are continuous everywhere, it implies that \(f(x)\) is continuous within the given interval. Continuity ensures that there are no breaks, jumps, or holes in the graph of the function, which is a crucial pre-requisite for applying Rolle’s Theorem.

In conclusion, our function \(f(x) = |9 - 4x^2|\), is continuous on the interval \(\bigg[-\frac{3}{2}, \frac{3}{2}\bigg]\). This should be a reassuring checkmark in our process of verifying Rolle’s Theorem.

Differentiability

Next, let’s dive into differentiability. Differentiability means that a function has a derivative at each point within the interval. This derivative represents the function's slope at each point. For a function to be differentiable, its graph must be smooth (no sharp bends or cusps).
However, for \(f(x) = |9 - 4x^2|\), the absolute value introduces potential problems. Specifically, the point \(x = 0\) deserves closer inspection.
At \(x = 0\), the function transitions from one parabola to another with different slopes. This creates a sharp point or a 'cusp,' where the function is not differentiable. Hence, while \(f(x)\) is differentiable on most of the interval, it fails at \(x = 0\).
In summary, since our function is not differentiable at \(x = 0\) due to the cusp, we conclude that \(f(x)\) is *not* differentiable on the entire open interval \(\bigg(-\frac{3}{2}, \frac{3}{2}\bigg)\).

Absolute Value Function

The absolute value function is key to our problem. Understanding it can help us see why Rolle’s Theorem may not apply here. The absolute value of a number is its distance from zero on the number line, regardless of direction. In mathematical terms, \(|x|\) equals \(x\) if \(x \geq 0\) and \(-x\) if \(x < 0\).
When this idea extends to functions, like \(f(x) = |9 - 4x^2|\), we get a piecewise function:

• For \(9 - 4x^2 \geq 0\), the function is simply \(9 - 4x^2\).
• For \(9 - 4x^2 < 0\), the function becomes \(-(9 - 4x^2)\) or \(4x^2 - 9\).

This transformation creates a graph where the smooth parabola \(9 - 4x^2\) gets flipped at the points where \(9 - 4x^2 = 0\). Since \(x = \pm\frac{3}{2}\) make the original expression zero, these points mark significant transitions.
In conclusion, the absolute value function here causes a cusp at \(x = 0\), which disrupts differentiability. This cusp is why \(f(x)\) does not satisfy all conditions of Rolle’s Theorem, even if it passes the continuity and endpoint conditions.