Find all the critical points of the function 𝑓(𝑥) = 𝑥^ 25 − 𝑥^ 9

Question

Use the First and/or Second Derivative Test to determine whether each critical point is a local maximum, a local minimum, or neither.You do not need to identify any global extrema.

Answer 1

A Critical Point occurs where the derivative of a function is equal to 0 or undefined.

Our first step for approaching this problem is to take the derivative using the power rule:

The derivative of x²⁵ - x⁹ = 25x²⁴ - 9x⁸

We can now set the function equal to 0 and factor

25x²⁴ - 9x⁸ = x⁸(25x¹⁶ - 9) = x⁸(5x⁸ - 3)(5x⁸ + 3)

Set each of these three terms equal to 0 and solve for x to find critical points:

x⁸ = 0, x = 0

5x⁸ - 3 = 0, 5x⁸ = 3, x⁸ = 3/5, x = ^8√3/5

5x⁸ + 3 = 0, 5x⁸ = -3, x⁸ = -3/5, x = - ^8√3/5

To determine whether these points are maximums or minimums, we have to look at the behavior of the function just before and after the point. We will choose points right before the actual function values and evaluate the derivative at those points.

When x = -1/2 the derivative equals a negative value. The slope is downward at this point. When x = 1/2 the derivative is a negative value. This means that x = 0 is neither a minimum nor a maximum because the behavior of the function doesn't change near the point.

When x = 4/5 the derivative is a negative value. When x = 1, the derivative is a positive value. The slope of the function goes from decreasing to increasing, so x = ^8√3/5 is a minimum.

When x = -1, the derivative is a positive value. When x = -4/5, the derivative is a negative value. The slope of the function goes from increasing to decreasing, so x = - ^8√3/5 is a maximum.

Answer 2

x^25 - x^9 = x^9(x^16 - 1) = x^9(x^4-1)(x^4+1) = x^9(x^2-1)(x^2+1)(x^4+1) = x(x-1)(x+1)(x^2+1)(x^4+1)

x = 0, 1, -1

0 has multiplicity 9, 1 and -1 just multiplicity 1. there are also 14 imaginary solutions (25-9-1-1)

the graph intersects the x axis 3 times, at x-0, 1, and -1 or (0,0), (1,0) and (-1,0)

f'(x) = 25x^24 -9x^8 = 0

x^8(25x^16 -9) = 0

x^8(5x^8-3)(5x^8+3) = 0

x=0 and x = (3/5)^1/8 = the 8th root of 0.6 = 0.938 or -0.938

the curve has zero slope at x=0 and almost 1 and almost -1

f'(0.938) =0 (0.938,-0.3602) and (-0.938, +0.3602) are local extrema, 1st point is a local max, 2nd local min

f'(0,-9.38) =0

f'(0) =0 (0,0) is an inflection point

f'(1) = 25-9 = 16 curve is upward sloping

f'(-1) = 25-9 = 16 curve is upward sloping at x=-1

f''(x) = 800x^23 - 72x^7

f''(0) =0 (0,0) is an inflection point where concavity changes, where slope changes sign

f"(1) >0 curve is concave up at x=1

f"(-1)<0 curve is concave down at x=-1

there are no global extrema, just two local extrema, max at (0,-.938) min at (0,.938)