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Determine the inputs that yield the minimum values for each function. Compute the minimum value in each case. (a) \(f(x)=\sqrt{x^{2}-6 x+73}\) (b) \(g(x)=\sqrt[3]{x^{2}-6 x+73}\) (c) \(h(x)=x^{4}-6 x^{2}+73\)

Short Answer

Expert verified
(a) Minimum is 8 at x = 3. (b) Minimum is 4 at x = 3. (c) Minimum is 82 at x = ±√3.

Step by step solution

01

Analyze the Function f(x)

The function given is \( f(x)=\sqrt{x^{2}-6x+73} \). The term \( x^2 - 6x + 73 \) is always non-negative since it is under a square root. To find the minimum value of \( f(x) \), we minimize \( x^2 - 6x + 73 \).
02

Find Critical Points for f(x)

Complete the square for \( x^2 - 6x + 73 \): \[x^2 - 6x + 9 + 64 = (x-3)^2 + 64\]The lowest value of \( (x-3)^2 \) is 0 when \( x = 3 \). Thus, the minimum value of \( f(x) \) is \( \sqrt{64} = 8 \) when \( x = 3 \).
03

Analyze the Function g(x)

The function given is \( g(x)=\sqrt[3]{x^{2}-6x+73} \). Similar to the previous function, we need to minimize \( x^2 - 6x + 73 \). The minimized expression is already obtained as \( (x-3)^2 + 64 \) with the lowest value of 64.
04

Compute Minimum Value for g(x)

Since \( g(x) = \sqrt[3]{x^{2}-6x+73} \) and the minimum inside the cube root is 64, the minimum value of \( g(x) \) is \( \sqrt[3]{64} = 4 \) when \( x = 3 \).
05

Analyze and Derivative of h(x)

The function is \( h(x)=x^4-6x^2+73 \). To find the critical points, take its derivative: \[h'(x) = 4x^3 - 12x\]Set the derivative to zero: \[4x^3 - 12x = 0 \]\[4x(x^2 - 3) = 0\]This gives \( x = 0 \), \( x = \sqrt{3} \), and \( x = -\sqrt{3} \).
06

Second Derivative Test for h(x)

To confirm whether these critical points are minima, calculate \( h''(x) = 12x^2 - 12 \). Evaluate at critical points:- \( h''(0) = -12 \) (indicating a concave down point at \( x = 0 \)), not a min.- \( h''(\sqrt{3}) = 24 \) (indicating a local min at \( x = \sqrt{3} \)).- \( h''(-\sqrt{3}) = 24 \) (indicating a local min at \( x = -\sqrt{3} \)).
07

Compute Minimum Value for h(x)

Calculate the value of \( h(x) \) at \( x = \sqrt{3} \) and \( x = -\sqrt{3} \):\[h(\sqrt{3}) = (\sqrt{3})^4 - 6(\sqrt{3})^2 + 73 = 27 - 18 + 73 = 82\]The same result is obtained at \( x = -\sqrt{3} \). Hence, the minimum value of \( h(x) \) is 82.

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Key Concepts

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

Critical Points
Critical points in calculus are essential when analyzing functions to find their extreme values, such as minima or maxima. They occur where the derivative of a function is either zero or undefined. Knowing this helps locate potential points where a function might be at a minimum or maximum.

For a function like here, critical points are particularly important as they guide us to the points where the behavior of the function changes. To find these points, you compute the derivative, set it to zero, and solve for the variable. For example, in the case of the function \( h(x) = x^4 - 6x^2 + 73 \), the derivative is \( h'(x) = 4x^3 - 12x \). By setting \( h'(x) = 0 \), you identify the critical points at \( x = 0, \sqrt{3}, \text{and} -\sqrt{3} \).

Understanding critical points lays the groundwork for further analysis like applying the derivative test to classify each critical point and understand their behavior in the function's graph.
Derivative Test
The derivative test is a key tool in analyzing the nature of critical points, helping us determine whether they represent local minima, maxima, or neither. Once you have identified the critical points, the next step is to check their type using the second derivative test.

This test involves taking the second derivative of the function. For instance, with \( h(x) = x^4 - 6x^2 + 73 \), its second derivative is \( 12x^2 - 12 \). By substituting the critical points into this second derivative, we can determine the concavity of the function at those points. If the second derivative at a critical point is positive, it indicates a local minimum (the graph is concave upwards). Conversely, if it's negative, it indicates a local maximum. If it is zero, the test is inconclusive.

In this situation, at \( x = \sqrt{3} \) and \( x = -\sqrt{3} \), \( h''(x) = 24 \), indicating these points are local minima. Meanwhile, at \( x = 0 \), the function is concave down with \( h''(0) = -12 \), suggesting a local maximum.
Function Analysis
Function analysis goes beyond just computing critical points or using derivative tests. It involves a complete understanding of the function’s behavior over its entire domain. Beginning with recognizing the type of function and its domain, you consider the long-term behavior, such as limits as \( x \) approaches positive or negative infinity, which helps ascertain where the function is boundless.

Then, analyze the symmetry or periodicity of the function, if applicable, which can reduce the effort needed in finding minima and maxima. Apply algebraic techniques like completing the square for functions involving squares or cubes to transform them into a simpler form that reveals intuitive insights about their behavior. For example, by completing the square for \( x^2 - 6x + 73 \) to \( (x-3)^2 + 64 \), it simplifies the process of analyzing the functions \( f(x) = \sqrt{x^2 - 6x + 73} \) and \( g(x) = \sqrt[3]{x^2 - 6x + 73} \), thereby facilitating the identification of minimum values.

In conclusion, function analysis incorporates these various tools and techniques to provide a holistic view of the function, making it easier to locate and confirm minima or maxima efficiently.

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