To find the relative maximum point of the function [tex]\( f(x) = \frac{x^3}{3} - \frac{x^2}{2} + 5 \)[/tex], we need to follow these steps:
### Step 1: Find the first derivative [tex]\( f'(x) \)[/tex]
The first derivative of the function will help us find the critical points.
[tex]\[ f(x) = \frac{x^3}{3} - \frac{x^2}{2} + 5 \][/tex]
To find the first derivative, we differentiate [tex]\( f(x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx} \left( \frac{x^3}{3} - \frac{x^2}{2} + 5 \right) = x^2 - x \][/tex]
### Step 2: Find the critical points
Critical points occur where the first derivative is zero or undefined. Here, we only need to set the first derivative to zero, as it is defined everywhere for this polynomial function.
[tex]\[ f'(x) = x^2 - x = 0 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ x(x - 1) = 0 \][/tex]
This gives us two critical points:
[tex]\[ x = 0 \][/tex]
[tex]\[ x = 1 \][/tex]
### Step 3: Find the second derivative [tex]\( f''(x) \)[/tex]
The second derivative will help us determine the concavity of the function at the critical points.
[tex]\[ f'(x) = x^2 - x \][/tex]
Differentiate again to find the second derivative:
[tex]\[ f''(x) = \frac{d}{dx} (x^2 - x) = 2x - 1 \][/tex]
### Step 4: Determine the concavity at each critical point
Evaluate the second derivative at each critical point to determine if it's a maximum or minimum.
For [tex]\( x = 0 \)[/tex]:
[tex]\[ f''(0) = 2(0) - 1 = -1 \][/tex]
Since [tex]\( f''(0) < 0 \)[/tex], the function is concave down at [tex]\( x = 0 \)[/tex], indicating a relative maximum.
For [tex]\( x = 1 \)[/tex]:
[tex]\[ f''(1) = 2(1) - 1 = 1 \][/tex]
Since [tex]\( f''(1) > 0 \)[/tex], the function is concave up at [tex]\( x = 1 \)[/tex], indicating a relative minimum.
### Step 5: Find the function value at the relative maximum point
To find the [tex]\( y \)[/tex]-coordinate of the relative maximum, substitute [tex]\( x = 0 \)[/tex] back into the original function [tex]\( f(x) \)[/tex]:
[tex]\[ f(0) = \frac{0^3}{3} - \frac{0^2}{2} + 5 = 5 \][/tex]
Thus, the relative maximum point of the function is:
[tex]\[ (0, 5) \][/tex]
Therefore, the correct answer is:
(D) [tex]\((0, 5)\)[/tex]
