Problem 19 - Applied Calculus

Question

For the functions in problems, do the following: (a) Find \(f^{\prime}\) and \(f^{\prime \prime}\). (b) Find the critical points of \(f\). (c) Find any inflection points of \(f\). (d) Evaluate \(f\) at its critical points and at the endpoints of the given interval. Identify local and global maxima and minima of \(f\) in the interval. (e) Graph \(f\). $$ f(x)=2 x^{3}-9 x^{2}+12 x+1(-0.5 \leq x \leq 3) $$

Step-by-Step Solution

Answer

Local max at \(x = 1\), \(f(1)=6\); global min at \(x = -0.5\), \(f(-0.5)=-5.875\).

Step 1: Differentiate to Find First and Second Derivatives

To find the first derivative \( f'(x) \), apply the power rule to each term of \( f(x) = 2x^3 - 9x^2 + 12x + 1 \). This yields \( f'(x) = 6x^2 - 18x + 12 \).Next, find the second derivative \( f''(x) \) by differentiating \( f'(x) \):\[ f''(x) = 12x - 18 \].Thus, \( f'(x) = 6x^2 - 18x + 12 \) and \( f''(x) = 12x - 18 \).

Step 2: Find Critical Points

To find critical points, set \( f'(x) = 0 \):\[ 6x^2 - 18x + 12 = 0 \].Divide the equation by 6:\[ x^2 - 3x + 2 = 0 \].Factor the quadratic equation:\[ (x-1)(x-2) = 0 \].Solving, we get the critical points \( x = 1 \) and \( x = 2 \).

Step 3: Find Inflection Points

To find inflection points, set \( f''(x) = 0 \):\[ 12x - 18 = 0 \].Solving for \( x \), we get:\[ x = 1.5 \].The point \( x = 1.5 \) is a potential inflection point. To confirm, check the sign changes of \( f''(x) \) around \( x = 1.5 \). Since \( f''(x) \) changes sign, there is an inflection point at \( x = 1.5 \).

Step 4: Evaluate Function at Critical Points and Endpoints

Evaluate the function \( f(x) \) at the critical points \( x = 1 \) and \( x = 2 \), and at the interval endpoints \( x = -0.5 \) and \( x = 3 \).Compute:* \( f(-0.5) = 2(-0.5)^3 - 9(-0.5)^2 + 12(-0.5) + 1 = -5.875 \)* \( f(1) = 2(1)^3 - 9(1)^2 + 12(1) + 1 = 6 \)* \( f(2) = 2(2)^3 - 9(2)^2 + 12(2) + 1 = 3 \)* \( f(3) = 2(3)^3 - 9(3)^2 + 12(3) + 1 = 1 \)The local maximum is at \( x = 1 \) with \( f(1) = 6 \), which is also the global maximum in the interval. The endpoint \( x = -0.5 \) gives the global minimum with \( f(-0.5) = -5.875 \).

Step 5: Sketch the Graph of the Function

Using the critical points \( x = 1 \) and \( x = 2 \), inflection point \( x = 1.5 \), evaluations from the critical points and endpoints, sketch the graph of \( f(x) \) on \( -0.5 \leq x \leq 3 \). Identify visible changes in concavity and confirm that \( x = 1.5 \) marks the inflection point. The function should show a local and global maximum at \( x = 1 \) and slightly decreases towards both ends of the interval.

Key Concepts

Critical Points

Critical points of a function are values of \( x \) where the derivative \( f'(x) \) equals zero or is undefined. These points are essential for identifying where a function's graph might change direction. To find them, solve \( f'(x) = 0 \). In our example, this involves setting \( 6x^2 - 18x + 12 = 0 \), which factors into \( (x-1)(x-2) = 0 \). Hence, the critical points are \( x = 1 \) and \( x = 2 \).
Critical points are where potential local maxima and minima occur. By evaluating the function at these points, and analyzing the sign of the derivative before and after these points, we can identify such critical characteristics of the graph.

Inflection Points

Inflection points occur where a function's graph changes concavity. This means the graph shifts from curving upwards to downwards or vice versa. To detect these points, set the second derivative \( f''(x) = 0 \) and look for sign changes around these solutions. In our example, solving \( f''(x) = 12x - 18 = 0 \) gives \( x = 1.5 \).
It's important to verify if \( f''(x) \) actually changes sign at \( x = 1.5 \). If there is a sign change, then the graph does indeed have an inflection point at this location. This indicates that point \( x = 1.5 \) is where the curve changes its concave behavior.

Function Derivatives

Derivatives are the backbone of calculus, helping us understand how functions change. Specifically, the first derivative \( f'(x) \) gives us the slope of the tangent line at any point, which tells us the rate of change of the function. For \( f(x) = 2x^3 - 9x^2 + 12x + 1 \), the first derivative is \( f'(x) = 6x^2 - 18x + 12 \). This derivative helps find critical points and reveal intervals where the function is increasing or decreasing.
The second derivative \( f''(x) \) provides insights into the concavity of the function, indicating where the function curves upwards or downwards. In this case, \( f''(x) = 12x - 18 \). Where \( f''(x) > 0 \), the function is concave up, and where \( f''(x) < 0 \), it is concave down. Understanding these derivatives is key to analyzing and graphing functions accurately.

Graph of a Function

Graphing a function is like drawing a map of its behavior using derivatives, critical points, and inflection points as clues. For \( f(x) = 2x^3 - 9x^2 + 12x + 1 \), identify critical points (\( x = 1 \) and \( x = 2 \)) and inflection point (\( x = 1.5 \)). This helps sketch how the curve rises or falls.
Evaluate function values at critical points and endpoints to determine the highest and lowest points in the interval \([-0.5, 3]\). The function hits a global maximum of 6 at \( x = 1 \) and a minimum of -5.875 at \( x = -0.5 \).

• The graph starts curving upwards from \(-0.5\), peaks at \(x = 1\), flattens at \(x = 2\), and finally gently curves towards \( x = 3 \).
• Inflection point at \(x = 1.5\) confirms a change in concavity.

This detailed analysis helps create a helpful and informative visualization of function behavior over its domain.