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Chapter 4: Problem 21

Find the value(s) of \(x\) that give critical points of \(y=\) \(a x^{2}+b x+c,\) where \(a, b, c\) are constants. Under what conditions on \(a, b, c\) is the critical value a maximum? A minimum?

Short Answer

Expert verified
The critical point is at \(x = -\frac{b}{2a}\); it's a minimum if \(a > 0\) and a maximum if \(a < 0\).

Step by step solution

01

Understand Critical Points

Critical points of a function occur where the derivative is zero or undefined. Since the function \(y = ax^2 + bx + c\) is a polynomial, its derivative is defined everywhere. We will find the derivative and set it equal to zero to find critical points.
02

Compute the Derivative

The function is a quadratic function, \(y = ax^2 + bx + c\). The derivative with respect to \(x\) is computed as follows:\[ \frac{dy}{dx} = 2ax + b \]
03

Find Critical Points

Set the derivative \(2ax + b\) equal to zero to find the critical points:\[ 2ax + b = 0 \]Solve for \(x\):\[ x = -\frac{b}{2a} \]This value of \(x\) is the critical point of the function.
04

Determine the Nature of the Critical Point

To determine whether the critical point is a maximum or minimum, we use the second derivative test. Compute the second derivative of the function:\[ \frac{d^2y}{dx^2} = 2a \]If \(2a > 0\), the critical point is a minimum because the parabola opens upwards. If \(2a < 0\), the critical point is a maximum because the parabola opens downwards.

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Key Concepts

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

Quadratic Functions
Quadratic functions are fundamental in calculus and algebra. These functions are usually expressed in the form \(y = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants and \(a eq 0\). A quadratic function forms a parabolic graph when plotted, which might either open upwards or downwards.

  • If the coefficient \(a > 0\), the parabola opens upwards, resembling a U-shape.
  • If \(a < 0\), the parabola opens downwards, resembling an upside-down U.

Quadratic functions are key because their simple structure allows us to easily find critical points, which help identify locations on the graph where the function may exhibit peaks (maxima) or troughs (minima).

These critical points are where the nature of the function's increase or decrease changes, typically at the vertex of the parabola.
Derivative Test
The derivative test is a method used to identify the critical points of a function, which are points where the function changes behavior from increasing to decreasing or vice versa. To find these, we start by taking the first derivative of the function. In our quadratic function \( y = ax^2 + bx + c \), the first derivative is \( \frac{dy}{dx} = 2ax + b \).

Setting the first derivative equal to zero identifies potential critical points, as it represents locations where the slope of the tangent is zero, indicating the potential peaks or troughs of the graph. Solving \(2ax + b = 0\) gives the critical point \(x = -\frac{b}{2a}\).

  • When \(\frac{dy}{dx} = 0\), it suggests the potential location of maxima or minima, but alone it doesn't provide the nature.
  • Additionally, if the derivative is undefined at some points, those would be checked as potential critical points as well, though not applicable in our polynomial case.
Second Derivative
The second derivative test further investigates the nature of the identified critical point(s) from the first derivative test. It effectively tells us whether we have a maximum, minimum, or an inflection point at these critical locations. To apply this test, we compute the second derivative of the function. For our quadratic function \( y = ax^2 + bx + c \), the second derivative is a constant, given by \( \frac{d^2y}{dx^2} = 2a \).

  • If \(2a > 0\), the critical point is a local minimum. This indicates that the parabola opens upwards.
  • Conversely, if \(2a < 0\), the critical point is a local maximum, indicating the parabola opens downwards.
  • If \(2a = 0\), the second derivative test is inconclusive, but this scenario does not occur for a valid quadratic function since \(a eq 0\).

Thus, the sign of \(2a\) fully determines the nature of the critical points on a parabolic graph, making the second derivative test a powerful tool in analyzing quadratic functions.

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Most popular questions from this chapter

If \(t\) is in hours, the drug concentration curve for a drug is given by \(C=17.2 t e^{-0.4 t} \mathrm{ng} / \mathrm{ml} .\) The minimum effective concentration is \(10 \mathrm{ng} / \mathrm{ml}\). (a) If the second dose of the drug is to be administered when the first dose becomes incffective, when should the second dose be given? (b) If you want the onset of effectiveness of the second dose to coincide with termination of effectiveness of the first dose, when should the second dose be given?

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