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Chapter 14: Problem 77

A function \(f\) is such that \(f^{\prime}(4)=f^{\prime \prime}(4)=0\) and \(f\) has minimum value 10 at \(x=4\). Then \(f(x)=\) (A) \(4+(x-4)^{4}\) (B) \(10+(x-4)^{4}\) (C) \((x-4)^{4}\) (D) None of these

Short Answer

Expert verified
(B) \(10 + (x - 4)^4\)

Step by step solution

01

Understanding the condition of derivatives

We are given that \(f^{\prime}(4) = f^{\prime \prime}(4) = 0\). This means that at \(x = 4\), the first and second derivatives of the function are both zero. To have a local minimum at this point, usually, the second derivative should be positive, but since it is zero, we need to analyze further or find a higher derivative.
02

Analyze possible functions

The function that checks the properties \(f^{\prime}(4) = 0\) and \(f^{\prime \prime}(4) = 0\), while having a minimum at \(x = 4\), could be of the form \(f(x) = a(x-4)^n + c\), where \(n > 2\) and we desire \(f(4) = 10\). For \(n = 4\), the third and fourth derivatives are needed to decide the concavity.
03

Substituting specific forms into conditions

Suppose \(f(x) = (x-4)^4 + c\). Then \(f(4) = 0^4 + c = c\). Given the minimum value is 10 at \(x = 4\), we must have \(f(4) = 10\). Thus, \(c = 10\), showing an expression \(f(x) = (x-4)^4 + 10\).
04

Verifying specific functional form

Given \(f(x) = 10 + (x-4)^4\), verify \(f^{\prime}(4) = 0\) and \(f^{\prime \prime}(4) = 0\). The derivative, \(f^{\prime}(x) = 4(x-4)^3\) and \(f^{\prime \prime}(x) = 12(x-4)^2\), both evaluate to 0 at \(x = 4\), satisfying the conditions.

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Key Concepts

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

Optimization
Optimization is the process of finding the best possible solution or outcome in a given situation. In mathematics, specifically calculus, it often involves finding the maximum or minimum values of a function. This process relies heavily on understanding the nature of the function and its derivatives.
  • By setting the first derivative equal to zero, we identify critical points, which indicate where potential maxima or minima might occur.
  • The second derivative can tell us more about the concavity of these points, helping determine if they are maxima, minima, or points of inflection.
The problem provided is an excellent example of using optimization techniques to determine the minimum value a function might have at a specific point. It is essential to understand the higher-order derivatives, especially when the first and second derivatives are zero, as they help shed light on the behavior of the function at those critical junctures.
Derivatives
Derivatives are a fundamental concept in calculus that measure how a function changes as its input changes. The first derivative of a function gives the slope of the tangent line at any point, telling us whether the function is increasing or decreasing at that point.
  • If the first derivative is zero, the function has a horizontal tangent at that point, suggesting a potential maximum, minimum, or saddle point.
  • The second derivative provides information on the concavity of the function. A positive second derivative indicates concavity upwards, suggesting a local minimum, while a negative one indicates concavity downwards, suggesting a local maximum.
In the exercise, both the first and second derivatives at the point 4 are zero, which presents a unique scenario. In such cases, higher-order derivatives become necessary to determine the function's behavior at that point.
Local Minimum
A local minimum of a function is a point where the function value is lower than at any nearby points. At a local minimum, the function changes direction from decreasing to increasing. For a student understanding this concept, it is essential to remember:
  • The first derivative is zero at a local minimum, indicating a potential change in direction.
  • In typical scenarios, the second derivative being positive confirms the presence of a local minimum.
However, the given exercise presents an atypical case where the second derivative is zero. This calls for examining higher-order derivatives. For a function to have a local minimum at such a point, we need the lowest non-zero derivative past the second to be positive. In the case of \[f(x) = 10 + (x-4)^4\], the fourth derivative would be non-zero and positive, confirming the local minimum at \(x = 4\) with a value of 10.

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Most popular questions from this chapter

The value of \(a\) in order that \(f(x)=\sin x-\cos x-a x+b\) decreases for all real values is given by (A) \(a \geq \sqrt{2}\) (B) \(a<\sqrt{2}\) (C) \(a \geq 1\) (D) \(a<1\)

The shortest distance between the line \(y-x=1\) and the curve \(x=y^{2}\) is (A) \(\frac{3 \sqrt{2}}{8}\) (B) \(\frac{2 \sqrt{3}}{8}\) (C) \(\frac{3 \sqrt{2}}{5}\) (D) \(\frac{\sqrt{3}}{4}\)

The difference between the greatest and least values of the function \(f(x)=\cos x+\frac{1}{2} \cos 2 x-\frac{1}{3} \cos 3 x\) is (A) \(2 / 3\) (B) \(8 / 7\) (C) \(9 / 4\) (D) \(3 / 8\)

If the function \(f(x)=2 x^{3}-9 a x^{2}+12 a^{2} x+1\), where \(a>0\), attains its maximum and minimum at \(p\) and \(q\) respectively such that \(p^{2}=q\), then \(a\) equals (A) 3 (B) 1 (C) 2 (D) \(\frac{1}{2}\)

If \(f(x)=\frac{x^{2}-1}{x^{2}+1}\), for every real number, then minimum value of \(\vec{f}\) (A) Does not exist (B) Is note attained even through \(f\) is bounded (C) Is equal to 1 (D) Is equal to \(-1\)
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