Answer :

The x-coordinates of the relative minima of f(x) = -(3/4)x⁴ + 14x³ - 60x² + 23 are 4 and 10, while the critical point at x = 0 corresponds to a relative maximum.

The relative minima of a function occur at points where the function reaches its lowest values within a certain interval. To find the x-coordinates of all relative minima of the function f(x) = -(3/4)x⁴ + 14x³ - 60x² + 23, we can follow these steps:

1. Calculate the derivative of f(x) with respect to x. The derivative gives us information about the slope of the function at different points.

2. Set the derivative equal to zero and solve for x to find the critical points. These are the points where the slope of the function is zero, which could correspond to relative minima, maxima, or points of inflection.

3. Determine the nature of each critical point by analyzing the second derivative. The second derivative helps us identify whether the critical point is a relative minimum, maximum, or neither.

Let's go through these steps in detail:

1. Calculate the derivative of f(x):
f'(x) = d/dx [-(3/4)x⁴ + 14x³ - 60x² + 23]

To find the derivative, we can apply the power rule and the sum rule of differentiation:
f'(x) = -3x³ + 42x² - 120x

2. Set the derivative equal to zero and solve for x to find the critical points:
-3x³ + 42x² - 120x = 0

Factoring out an x, we get:
x(-3x² + 42x - 120) = 0

Now, we have two possibilities:
a) x = 0 (which is always a critical point since it makes the derivative zero)
b) -3x² + 42x - 120 = 0

To solve the quadratic equation, we can factor or use the quadratic formula. Factoring gives us:
-3(x - 4)(x - 10) = 0

Therefore, the critical points are:
x = 0, x = 4, x = 10

3. Determine the nature of each critical point by analyzing the second derivative:
The second derivative, f''(x), can help us identify whether each critical point is a relative minimum, maximum, or neither.

f''(x) = d²/dx² [-3x³ + 42x² - 120x]

Using the power rule and the sum rule again, we get:
f''(x) = -9x² + 84x - 120

Now, substitute the critical points into f''(x):
For x = 0: f''(0) = -9(0)² + 84(0) - 120 = -120
For x = 4: f''(4) = -9(4)² + 84(4) - 120 = 0
For x = 10: f''(10) = -9(10)² + 84(10) - 120 = 0

By analyzing the signs of f''(x) at each critical point, we can determine their nature:
a) x = 0: Since f''(0) < 0, the critical point x = 0 corresponds to a relative maximum.
b) x = 4: Since f''(4) = 0, the nature of this critical point is inconclusive.
c) x = 10: Since f''(10) = 0, the nature of this critical point is also inconclusive.

In summary, the x-coordinates of the relative minima of f(x) = -(3/4)x⁴ + 14x³ - 60x² + 23 are 4 and 10, while the critical point at x = 0 corresponds to a relative maximum.

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