Problem 47 - Calculus Early Transcendentals

Question

a. Find the critical points of \(f\) on the given interval. b. Determine the absolute extreme values of \(f\) on the given interval when they exist. c. Use a graphing utility to confirm your conclusions. $$f(x)=2 x^{3}-15 x^{2}+24 x \text { on }[0,5]$$

Step-by-Step Solution

Answer

Answer: The absolute maximum value of \(f(x)\) is \(11\) at \(x = 1\), and the absolute minimum value is \(-16\) at \(x = 4\).

Step 1: Find the first derivative of the function

To find the critical points of the function, we must first find the first derivative of \(f(x)\). The first derivative of \(f(x)\) is given by\(f'(x) = \frac{d}{dx}(2x^3 - 15x^2 + 24x)\). Use the power rule to differentiate the given function: $$ f'(x) = 6x^{2}-30 x+24 $$

Step 2: Find the critical points

To find the critical points of \(f(x)\), set the first derivative equal to zero and solve for \(x\): $$ 6x^2 - 30x + 24 = 0 $$ We can simplify this quadratic equation by dividing by 6: $$ x^2 - 5x + 4 = 0 $$ Now, factor the quadratic equation: $$ (x - 4)(x - 1) = 0 $$ So the critical points are \(x = 1\) and \(x = 4\).

Step 3: Determine the extreme values of \(f(x)\) on the interval \([0,5]\)

To find the absolute extreme values of \(f(x)\), evaluate the function at the critical points and the endpoints of the interval: $$ f(0) = 2(0)^3 - 15(0)^2 + 24(0) = 0 \\ f(1) = 2(1)^3 - 15(1)^2 + 24(1) = 2 - 15 + 24 = 11 \\ f(4) = 2(4)^3 - 15(4)^2 + 24(4) = 128 - 240 + 96 = -16 \\ f(5) = 2(5)^3 - 15(5)^2 + 24(5) = 250 - 375 + 120 = -5 \\ $$ From these calculations, the absolute maximum value of \(f(x)\) is \(11\) at \(x = 1\), and the absolute minimum value is \(-16\) at \(x = 4\).

Step 4: Use a graphing utility to confirm the results

Using a graphing utility (such as Desmos or Geogebra), we can plot the function \(f(x) = 2x^3 - 15x^2 + 24x\) on the interval \([0,5]\). The graph confirms that there is a maximum value of \(11\) at \(x = 1\) and a minimum value of \(-16\) at \(x = 4\).

Key Concepts

First Derivative

The first derivative of a function is a foundational concept in calculus. When we calculate the derivative of a function like our given function, f(x) = 2x^3 - 15x^2 + 24x, we're finding a new function, f'(x), that shows us the rate at which f(x) is changing at any point x. To get the first derivative, we use rules of differentiation, like the power rule, which tell us that the derivative of x^n is nx^(n-1).

So, for our function, that means f'(x) = 6x^2 - 30x + 24. This first derivative is critical in finding where the graph of the function has horizontal tangents or changes direction, which brings us to the next section.

Absolute Extreme Values

Finding the absolute extreme values of a function within a given interval is like a treasure hunt for the highest and lowest points. These are points where the function reaches its maximum or minimum values, respectively. To locate these treasures, we first determine the critical points where the first derivative is zero or undefined, because these are possible locations of local extremes. Then, we evaluate the function at these critical points as well as at the endpoints of the interval—these are our potential 'x marks the spot’.

In our exercise, by plugging in the critical points and the endpoints into our original function f(x), we found that the highest treasure, the absolute maximum, is at x = 1 with a value of f(x) = 11, and the lowest, the absolute minimum, is at x = 4 with f(x) = -16.

Function Graphing

Graphing a function can turn complex algebraic expressions into a clearer visual representation, helping us understand the behavior of the function over an interval. The x-axis represents the input values, while the y-axis represents the output values calculated by the function. It's particularly helpful when we need to confirm findings analytically determined, like the extreme values and the shape of a curve.

By plotting our function on a graph, we can visualize the rise and fall of the curve between the interval [0,5], confirming our previously found maximum and minimum values. The points representing the extreme values, often referred to as the peaks and valleys on the graph, are where the curve changes direction.

Quadratic Equation Factoring

Factoring quadratic equations is a technique used to break them down into more manageable pieces. It's like dismantling a puzzle into its component pieces to understand the bigger picture. This process is particularly useful when we're trying to find the roots or the zero-points of a quadratic equation, which in turn helps us determine the critical points of a function.

To factor a quadratic equation, look for two numbers that multiply to give you the constant term (in our case, +4) and add up to the coefficient of the linear term (in our case, -5). Once factored, as we saw in our equation (x - 4)(x - 1) = 0, we can easily see that the roots are x=4 and x=1. These are the values where the function crosses the x-axis, and also our function's critical points.