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Chapter 3: Problem 14

In Exercises \(7-24\), sketch the graph of the function and find its absolute maximum and absolute minimum values, if any. $$ g(x)=2 x^{2}-3 x+1 \text { on }[0,1) $$

Short Answer

Expert verified
The graph of the function \(g(x)=2x^2-3x+1\) on the interval [0,1) is a parabola that opens upwards. The absolute maximum is 1 at \(x=0\) and the absolute minimum is \(\frac{1}{8}\) at \(x=\frac{3}{4}\).

Step by step solution

01

Determine critical points within the interval

To find the critical points of the function \(g(x)=2x^2-3x+1\), we need to take its derivative and find where the derivative is equal to 0 or undefined. In this case, the function is a polynomial, so the derivative will never be undefined. Derivative of \(g(x)\) is: \(g'(x)=4x-3\). Now we need to solve for \(g'(x)=0\): \(4x-3 = 0\) \(4x = 3\) \(x = \frac{3}{4}\) The function has one critical point at \(x=\frac{3}{4}\).
02

Evaluate the function at the critical points and endpoints

We will now evaluate \(g(x)\) at the critical point found in Step 1, and at the endpoints of the interval [0,1). Remember that the right endpoint is not included in the interval, so we'll calculate the limit as x approaches 1. \(g(0)=2(0)^2-3(0)+1=1\) \(g\left(\frac{3}{4}\right)=2\left(\frac{3}{4}\right)^2-3\left(\frac{3}{4}\right)+1=\frac{1}{8}\) Now we will find the limit as x approaches 1: $$ \lim_{x\to 1} g(x) = \lim_{x\to 1} (2x^2-3x+1) = 2(1)^2 - 3(1) +1 = 0 $$
03

Compare the function values to find the absolute maximum and absolute minimum

To find the absolute maximum and minimum, we will compare the function values obtained in Step 2: \(g(0)=1\) \(g\left(\frac{3}{4}\right)=\frac{1}{8}\) \(\lim_{x\to 1} g(x) = 0\) Since \(\frac{1}{8}\) is the smallest value, the absolute minimum of the function on the interval [0, 1) is \(\frac{1}{8}\) and occurs at \(x=\frac{3}{4}\). Since \(1\) is the largest value, the absolute maximum of the function on the interval [0, 1) is \(1\) and occurs at \(x=0\).
04

Sketch the graph of the function

To sketch the graph of the function on the interval [0, 1), use the information obtained in the previous steps: 1. The function \(g(x)=2x^2-3x+1\) is a parabola that opens upward. 2. The critical point \(x=\frac{3}{4}\) corresponds to the vertex of the parabola. 3. The absolute maximum occurs at x=0, and the function value is 1. 4. The absolute minimum occurs at \(x=\frac{3}{4}\), and the function value is \(\frac{1}{8}\). 5. The limit as x approaches 1 is 0. Based on these findings, sketch the parabola with the vertex at \(\left(\frac{3}{4},\frac{1}{8}\right)\), a point at (0,1), and the function value approaching 0 as x approaches 1. (Note: A graphing tool would be helpful to visualize the graph.)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Critical Points
Critical points are essential in analyzing a function's behavior, as they are where the derivative of the function is zero or undefined. For our function \(g(x)=2x^2-3x+1\), the critical points help identify locations where the function may achieve its maxima or minima.
To find the critical points, we first take the derivative of the function. The derivative, denoted as \(g'(x)\), gives us: \[g'(x) = 4x - 3\]
The next step is to solve \(g'(x) = 0\), which yields the equation:
  • \(4x - 3 = 0\)
  • \(x = \frac{3}{4}\)
Thus, the function has a critical point at \(x=\frac{3}{4}\). This point is vital for evaluating extreme function values on the given interval.
Derivative
The derivative is a fundamental concept in calculus that represents the rate of change of a function's output with respect to its input. For the function \(g(x) = 2x^2 - 3x + 1\), deriving it helps us determine the slope at any given point. The process involves finding the function's derivative:
  • The derivative of \(2x^2\) is \(4x\)
  • The derivative of \(-3x\) is \(-3\)
  • The derivative of the constant \(1\) is \(0\)
Thus, the complete derivative of the function is:\[g'(x) = 4x - 3\]
This derivative function is crucial for identifying critical points and understanding the overall shape of the graph.
Absolute Maximum and Minimum
Absolute maximum and minimum refer to the highest and lowest values, respectively, that a function achieves over a specific interval. For the function\(g(x)=2x^2-3x+1\) on the interval \([0,1)\), we calculate these at critical points and boundaries.
First, evaluate the function at the endpoints:
  • \(g(0)=1\)
Next, evaluate at the critical point:
  • \(g\left(\frac{3}{4}\right)=\frac{1}{8}\)
Finally, determine the limit as \(x\) approaches the right endpoint:
  • \(\lim_{x\to 1} g(x) = 0\)
Comparing these values, we find:
  • The absolute maximum is \(1\), occurring at \(x=0\)
  • The absolute minimum is \(\frac{1}{8}\), occurring at \(x=\frac{3}{4}\)
Parabola
A parabola is a U-shaped curve that is the graph of a quadratic function. For the function \(g(x) = 2x^2 - 3x + 1\), we have a parabola that opens upward, indicating it has a minimum point at its vertex.
The crucial factors about this parabola on the interval [0,1) include:
  • The function opens upward because the coefficient of \(x^2\) is positive (2).
  • The vertex, or turning point, is a critical aspect and is found at \(x=\frac{3}{4}\), corresponding to our critical point.
  • Knowing these traits helps in sketching the curve accurately.
The vertex forms the minimum point in this context, with the graph dipping at \(\left(\frac{3}{4},\frac{1}{8}\right)\) before rising again as \(x\) approaches higher values.
Limit
The limit examines the behavior of a function as the input approaches a particular point. For the function \(g(x) = 2x^2 - 3x + 1\), this is particularly important near the endpoints of the interval. Knowing limits allows us to assess function behavior as it nears but does not reach a boundary.
In this exercise, we calculate the limit as \(x\) approaches the right endpoint 1:\[\lim_{x\to 1} g(x) = 2(1)^2 - 3(1) + 1 = 0\]
  • This calculation provides the function value as \(x\) nears 1, confirming that \(g(x)\) approaches 0.
  • Limits enable us to infer continuous behavior across the interval.
  • Such information, combined with critical points, assists in accurately sketching the function's graph.

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