Questions: The graph of y=(x^2-2)^4(x^2+2)^5 is shown to the right. Find the coordinates of the local maxima and minima. Find the maximum points. (Type an ordered pair, using integers or decimals. Round the final answer to two decimal places as needed. Round all intermediate values to two decimal places as needed. Use a comma to separate answers, but do not use commas in any individual numbers.)

The graph of y=(x^2-2)^4(x^2+2)^5 is shown to the right. Find the coordinates of the local maxima and minima.

Find the maximum points.
(Type an ordered pair, using integers or decimals. Round the final answer to two decimal places as needed. Round all intermediate values to two decimal places as needed. Use a comma to separate answers, but do not use commas in any individual numbers.)

Transcript text: The graph of $y=\left(x^{2}-2\right)^{4}\left(x^{2}+2\right)^{5}$ is shown to the right. Find the coordinates of the local maxima and minima. Find the maximum points. $\square$ (Type an ordered pair, using integers or decimals. Round the final answer to two decimal places as needed. Round all intermediate values to two decimal places as needed. Use a comma to separate answers, but do not use commas in any individual numbers.)

Solution

Solution Steps

Step 1: Find the critical points

The given hint states that _dy/dx_ = 2_x_(x² - 2)³_(x² + 2)⁴_[9x² - 2]. To find the critical points, we set _dy/dx_ = 0 and solve for x. 2_x_(x² - 2)³_(x² + 2)⁴_[9x² - 2] = 0 This equation is satisfied when:

x = 0
x² - 2 = 0 => x = ±√2 => x ≈ ±1.41
x² + 2 = 0 (no real solutions)
9x² - 2 = 0 => x² = 2/9 => x = ±√(2/9) => x ≈ ±0.47

Step 2: Identify the maximum points

From the graph, we can observe a local maximum occurs near x = 0 and two other local maxima occur roughly symmetric about the y-axis, close to ±0.5. Comparing this to the critical points in Step 1, the local maxima occur at x ≈ ±0.47. Now, we substitute x = ±√(2/9) into the original equation y = (x² - 2)⁴*(x² + 2)⁵ to find the y-coordinates:

y = ((2/9) - 2)⁴_((2/9) + 2)⁵ y = (-16/9)⁴_(20/9)⁵ y ≈ (5.345)*(39.065) y ≈ 208.84

So, the maximum points are approximately (-0.47, 208.84) and (0.47, 208.84).