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Show that \(f(x)=-\left|x^{2}-4\right|\) has local maxima at \(x=2\) and \(x=-2\) but \(f(x)\) is not differentiable at \(x=2\) or \(x=-2\).

Short Answer

Expert verified
Local maxima occur at \(x=2\) and \(x=-2\) because of slope changes, but \(f(x)\) isn't differentiable there due to a jump in derivative values.

Step by step solution

01

Understanding the Function

The function in question is given by \( f(x) = -\left|x^2 - 4\right| \). To find the behavior at critical points like \(x=2\) and \(x=-2\), we need to consider how the absolute value function behaves and affects \(f(x)\).
02

Splitting Function Based on Absolute Value

The term \( |x^2 - 4| \) changes its formula based on whether \( x^2 - 4 \) is non-negative or negative. So for \(x^2 - 4 \geq 0\), \( |x^2 - 4| = x^2 - 4 \), and for \(x^2 - 4 < 0\), \(|x^2 - 4| = -(x^2 - 4)\).
03

Evaluating the Regions Separately

For \( x^2 < 4 \), \( f(x) = -(-(x^2 - 4)) = x^2 - 4 \). For \( x^2 \geq 4 \), \( f(x) = -(x^2 - 4) \).
04

Checking Local Maxima At Different Regions

Let's check behaviors around \( x=2 \) and \( x=-2 \). Consider points: \(x = 2\) and very slightly less and more than 2: \( f(2) = -|0| = 0\), \(f(2- ext{small}) \approx 2^2 - 4 = 0\), \( f(2+ ext{small}) \approx -(2.1^2 - 4) < 0\), showing a drop from \( x = 2 \). Similarly for \(x=-2\), as \(f(-2) = 0\), the surrounding values follow the same pattern. Thus, these are local maxima.
05

Check Differentiability

To check differentiability, consider the slope of \(f(x)\) from both sides at \(x=2\). Just before \(x=2\), the formula is \(x^2 - 4\) so the slope would be approximately \(2x\) when \(x=2\), giving 4. Just after \(x=2\), the formula changes and derivative becomes approximately \(-2x\), giving -4. So, the derivatives have a discontinuous jump.
06

Conclusion

At both \(x = 2\) and \(x = -2\), the function \(f(x) = -|x^2 - 4|\) achieves local maximum values but due to the discontinuous nature of derivatives at these points, the function is not differentiable at \(x = 2\) and \(x = -2\).

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Key Concepts

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

Local Maxima
In mathematics, a local maximum of a function is a point where the function reaches a peak in its neighborhood, which means the function value is greater than or equal to nearby values. For a function to have a local maximum at a particular point, the function value at that point should be greater than or equal to the function values for other points close to it.

This occurs around critical points, where the first derivative of the function either equals zero or is not defined. Consider the function given in the problem:
  • \( f(x) = -|x^2 - 4| \).
At \( x = 2 \) and \( x = -2 \), the function behaves such that just before these points, the value stays the same as at the point, and just after, the value decreases. Thus, these points exhibit a local maximum characteristic.

A local maximum doesn't mean it is the highest point globally over the entire function, but rather it simply is the highest point over a small interval.
Absolute Value Function
The absolute value function, denoted as \( |x| \), returns the non-negative value of a given real number \( x \). This implies that if \( x \) is negative, the absolute value function will multiply it by -1 to make it positive. This is important as it affects how functions that include absolute values behave.

For instance, in the function \( f(x) = -|x^2 - 4| \):
  • If \( x^2 - 4 \geq 0 \), then \( |x^2 - 4| = x^2 - 4 \), resulting in \( f(x) = -(x^2 - 4) \).
  • If \( x^2 - 4 < 0 \), then \( |x^2 - 4| = -(x^2 - 4) \), leading to \( f(x) = x^2 - 4 \).
This piecewise nature makes absolute value functions interesting as they can easily switch expressions across certain points, changing the function's behavior rapidly.

This abrupt change contributes to non-differentiability at certain critical points in the overall function.
Critical Points
Critical points of a function are crucial for analyzing its graphs and finding maxima and minima. They occur where the derivative of a function is zero or undefined. These points mark potential locations for local maxima, minima, or inflection points.

In the exercise function \( f(x) = -|x^2 - 4| \), the points \( x = 2 \) and \( x = -2 \) are critical because:
  • The derivative at these points doesn't exist due to the abrupt behavior change in the absolute value function.
  • Checking the behavior slightly before and after \( x = 2 \) shows a switch from an increasing to decreasing slope. This aligns \( x = 2 \) as a local maximum.
Discontinuities in derivatives, especially at such critical points, indicate the locations where the function isn't smooth and differentiable. Correspondingly, at critical points like \( x = 2 \) and \( x = -2 \), one can observe local maxima and nonsmooth behavior. Understanding how these points work helps in sketching accurate graphs and understanding function behavior.

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Most popular questions from this chapter

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