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Show that: x5 - 5x4 - 1 = 0 is

maximum when x-1, minimum when x-3, neither
when
x=0.

1 Answer

5 votes

Answer:

To determine whether the equation x^5 - 5x^4 - 1 = 0 has a maximum or minimum at certain values of x, we need to analyze the behavior of the function.

Let's take the derivative of the function and find the critical points:

f(x) = x^5 - 5x^4 - 1

f'(x) = 5x^4 - 20x^3

Setting f'(x) = 0 to find the critical points:

5x^4 - 20x^3 = 0

Factor out common terms:

5x^3(x - 4) = 0

This equation has two critical points:

1) x = 0

2) x = 4

To determine whether these points correspond to a maximum or minimum, we can use the second derivative test. Let's find the second derivative:

f''(x) = 20x^3 - 60x^2

Now, substitute the critical points into the second derivative:

1) For x = 0:

f''(0) = 20(0)^3 - 60(0)^2 = 0

Since the second derivative is zero, the test is inconclusive.

2) For x = 4:

f''(4) = 20(4)^3 - 60(4)^2 = 320

Since the second derivative is positive (320 > 0), this indicates a minimum at x = 4.

Based on the analysis, we find that the function has a minimum at x = 4. We also know that x = 0 is not a maximum or minimum point.

Therefore, the function x^5 - 5x^4 - 1 = 0 is minimum at x = 4, while x = 0 does not correspond to a maximum or minimum point.

User Sachsure
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