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Find the relative extrema using both first and second derivative tests. \(f(x)=1+8 x-3 x^{2}\)

Short Answer

Expert verified
The function has a local maximum at \( x = \frac{4}{3} \), with a value of \( \frac{25}{3} \).

Step by step solution

01

Calculate the First Derivative

Find the first derivative of the function \( f(x) = 1 + 8x - 3x^2 \). The derivative, \( f'(x) \), can be calculated using the power rule. For the term \( 8x \), the derivative is \( 8 \), and for \( -3x^2 \), it is \( -6x \). Therefore, \( f'(x) = 8 - 6x \).
02

Find Critical Points from the First Derivative

Set the first derivative \( f'(x) = 8 - 6x = 0 \) to find the critical points. Solving for \( x \), we get \( 8 = 6x \) which implies \( x = \frac{4}{3} \).
03

Use the Second Derivative for Concavity

Find the second derivative, \( f''(x) \), by differentiating \( f'(x) = 8 - 6x \). The derivative is \( f''(x) = -6 \). Since \( f''(x) \) is a constant negative value, this confirms concavity.
04

Apply the Second Derivative Test

Evaluate \( f''(x) \) at the critical point \( x = \frac{4}{3} \). Since \( f''( \frac{4}{3} ) = -6 \), which is negative, the function is concave down at this critical point, indicating a local maximum.
05

Verify Using the First Derivative Test

Check the sign of \( f'(x) \) to the left and right of \( x = \frac{4}{3} \). Choose a test point \( x = 1 \) (left of \( \frac{4}{3} \)) in \( f'(x) = 8 - 6x \), giving \( f'(1) = 2 \), positive. Choose \( x = 2 \) (right of \( \frac{4}{3} \)), which results in \( f'(2) = -4 \), negative. The sign change from positive to negative confirms a local maximum at \( x = \frac{4}{3} \).
06

Conclusion

The local maximum is at \( x = \frac{4}{3} \). Evaluating the original function at this point gives \( f(\frac{4}{3}) = 1 + 8 \times \frac{4}{3} - 3 \times (\frac{4}{3})^2 = \frac{25}{3} \). Hence, the local maximum point is \( \left( \frac{4}{3}, \frac{25}{3} \right) \).

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Key Concepts

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

First Derivative Test
The First Derivative Test helps to determine whether a critical point is a relative maximum, minimum, or neither. Begin by taking the first derivative of the function. For example, given the function \( f(x) = 1 + 8x - 3x^2 \), the first derivative is \( f'(x) = 8 - 6x \). Critical points occur where \( f'(x) = 0 \). Solving \( 8 - 6x = 0 \), we find \( x = \frac{4}{3} \), which is a critical point.Next, observe the sign change of \( f'(x) \) around this critical point:
  • If \( f'(x) \) changes from positive to negative, the critical point is a relative maximum.
  • If it changes from negative to positive, the critical point is a relative minimum.
For \( f(x) = 1 + 8x - 3x^2 \), the first derivative changes from positive to negative around \( x = \frac{4}{3} \), indicating a local maximum.
Second Derivative Test
The Second Derivative Test is a powerful tool to confirm the nature of a critical point found using the first derivative. It involves taking the second derivative of the original function.With \( f(x) = 1 + 8x - 3x^2 \), the second derivative is \( f''(x) = -6 \). Notice how it's a constant and negative, indicating the function is concave downward everywhere.The test involves evaluating the second derivative at the critical points found in the first derivative test:
  • If \( f''(x) > 0 \), it suggests the function is concave up, signifying a local minimum.
  • If \( f''(x) < 0 \), as in this example, it implies a concave down shape, confirming a local maximum.
Thus, for \( x = \frac{4}{3} \), since \( f''(\frac{4}{3}) = -6 \), we confirm a local maximum.
Critical Points
Critical points are the values of \( x \) where the first derivative is zero or undefined. These points are crucial in identifying the function's relative extrema, or local maximums and minimums.To find critical points from the function \( f(x) = 1 + 8x - 3x^2 \), first, calculate the derivative \( f'(x) = 8 - 6x \). Set \( f'(x) = 0 \) and solve for \( x \). The result is \( x = \frac{4}{3} \), the critical point of interest here.Knowing where critical points occur allows us to examine the behavior of the function around those points. By using first and second derivative tests, you can characterize these points to determine whether they correspond to local maxima, minima, or saddle points.

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Most popular questions from this chapter

If an unknown physical quantity \(x\) is measured \(n\) times, the measurements \(x_{1}, x_{2}, \ldots, x_{n}\) often vary because of uncontrollable factors such as temperature, atmospheric pressure, and so forth. Thus, a scientist is often faced with the problem of using \(n\) different observed measurements to obtain an estimate \(\bar{x}\) of an unknown quantity \(x\). One method for making such an estimate is based on the least squares principle, which states that the estimate \(\bar{x}\) should be chosen to minimize $$ s=\left(x_{1}-\bar{x}\right)^{2}+\left(x_{2}-\bar{x}\right)^{2}+\cdots+\left(x_{n}-\bar{x}\right)^{2} $$ which is the sum of the squares of the deviations between the estimate \(\bar{x}\) and the measured values. Show that the estimate resulting from the least squares principle is $$ \bar{x}=\frac{1}{n}\left(x_{1}+x_{2}+\cdots+x_{n}\right) $$ that is, \(\bar{x}\) is the arithmetic average of the observed values.

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