Question

$$ F(x)=-x^{4}+8 x^{2}+9 $$ (a) Determine whether \(F\) is even, odd, or neither. (b) There is a local maximum value of 25 at \(x=2 .\) Find a second local maximum value. (c) Suppose the area of the region enclosed by the graph of \(F\) and the \(x\) -axis between \(x=0\) and \(x=3\) is 50.4 square units. Using the result from (a), determine the area of the region enclosed by the graph of \(F\) and the \(x\) -axis between \(x=-3\) and \(x=0\).

Short answer

a) The function is even.b) The second local maximum is 25 at \(x = -2\).c) The area is 50.4 square units.

Step-by-step solution

Check if the function is even, odd, or neither

To determine if the function is even, odd, or neither, use the definitions:- A function is even if \(f(-x) = f(x)\).- A function is odd if \(f(-x) = -f(x)\).- If neither of these conditions is met, the function is neither.Evaluate \(F(-x)\):\[F(-x) = -(-x)^4 + 8(-x)^2 + 9 = -x^4 + 8x^2 + 9\]Since \(F(-x) = F(x)\), the function \(F(x)\) is even.

Find the second local maximum value

Given that there is a local maximum value of 25 at \(x=2\), find another local maximum by differentiating the function to find critical points:\[F'(x) = \frac{d}{dx}(-x^4 + 8 x^2 + 9) = -4x^3 + 16x\]Set the derivative equal to zero and solve for \(x\):\(-4x^3 + 16x = 0\)Factor out the common term:\[x(-4x^2 + 16) = 0\]\[x = 0\] or \(-4x^2 + 16 = 0\Rightarrow x^2 = 4\Rightarrow x = 2\) or \(x = -2\)Since \(x = 2\) is given as a local maximum, evaluate \(F(-2)\):\[F(-2) = -(-2)^4 + 8(-2)^2 + 9 = -16 + 32 + 9 = 25\]Thus, there is also a local maximum value of 25 at \(x = -2\).

Determine the area of the region enclosed between \(x = -3\) and \(x = 0\)

Since the function \(F(x)\) is even, the area between \(x = -a\) and \(x = a\) is twice the area between \(x = 0\) and \(x = a\):Given that the area between \(x = 0\) and \(x = 3\) is 50.4 square units, the area from \(x = -3\) to \(x = 3\) is:\[2 \times 50.4 = 100.8 \text{ square units}\]Hence, the area between \(x = -3\) and \(x = 0\) is also 50.4 square units.

Key concepts

even function

The concept of an even function is fundamental in mathematics, especially when dealing with polynomial functions like the one in this exercise. An even function exhibits a specific type of symmetric behavior with respect to the y-axis. To determine if a function is even, we need to check whether it satisfies the condition: \[ f(-x) = f(x) \]
For instance, in this exercise, our function is given as:\[ F(x) = -x^4 + 8x^2 + 9 \]Evaluating this at \( -x \), we get:\[ F(-x) = -(-x)^4 + 8(-x)^2 + 9 \]Simplifying this, we observe that:
\[ -x^4 + 8x^2 + 9 = F(x) \]Thus, \( F(x) \) is an even function. This symmetrical property can help in various calculations, as it allows us to understand that the function behaves identically on both sides of the y-axis.
We use this property later to determine enclosed areas for integration, simplifying the task by considering symmetry.

• An even function means f(x) behaves the same way for positive and negative values of x.
• The graph of an even function is symmetric about the y-axis.

local maximum

Understanding local maximum values is crucial in the study of calculus, particularly for graphing and optimization problems. A local maximum of a function is where the function value is higher than at all nearby points. It provides important information about the function's behavior. In this case, we already know a local maximum at \( x = 2 \). To find others, we use the derivative: \[ F'(x) = \frac{d}{dx}(-x^4 + 8x^2 + 9) = -4x^3 + 16x \]Setting \( F'(x) = 0 \) to find critical points: \[ -4x^3 + 16x = 0 \]we factorize and solve it:\[ x(-4x^2 + 16) = 0 \]This gives us: \[ x = 0 \text{ or } \ x^2 = 4 \rightarrow x = 2 \text{ or } x = -2 \]Since \( x = 2 \) is a known local maximum, evaluate \( F(-2) \):\[ F(-2) = -(-2)^4 + 8(-2)^2 + 9 = -16 + 32 + 9 = 25 \]It confirms another local maximum at \( x = -2 \) with the same value 25.

• A local maximum is where the function reaches a peak within a given range.
• It provides critical points essential for understanding the function’s highest values in specific intervals.

definite integral

The definite integral is a powerful tool for calculating the area under a curve. It’s essential for comprehending areas enclosed by graphs, as in this exercise. A definite integral from \( a \) to \( b \) of a function tells us the net area between the curve and the x-axis over that interval. Here, we were given the area from \( x = 0 \) to \( x = 3 \) is 50.4 square units.Since \( F(x) \) is known to be even:

• The area from \( x = -3 \) to \( x = 0 \) will be identical to the area from \( x = 0 \) to \( x = 3 \) because of symmetry.
• Hence, the total area from \( x = -3 \) to \( x = 3 \) is twice the given area.

Mathematically, this is:\[ 2 \times 50.4 = 100.8 \text{ square units.} \]Thus, understanding the symmetry involved simplified our problem!

• Definite integrals help measure the total area under a curve.
• Properties of even functions simplify integral calculations by leveraging symmetry.