The graph of function `f(x)=(x^(5))/(20)-(x^(4))/(12)+5` has :

Options

A. no local extremum one point of inflection.
B. two local maximum, one local minimum, two point of inflection
C. one local maximum, one local minimum, one point of inflection.
D. one local maximum, one local minimum, two point of inflection.

Explanation

To analyze the function \( f(x) = \frac{x^5}{20} - \frac{x^4}{12} + 5 \) for local maxima, local minima, and points of inflection, we will follow these steps:

### Step 1: Find the first derivative \( f'(x) \)
To find the critical points, we first need to differentiate the function.

\[ f'(x) = \frac{d}{dx}\left(\frac{x^5}{20}\right) - \frac{d}{dx}\left(\frac{x^4}{12}\right) + \frac{d}{dx}(5) \]

Using the power rule \( \frac{d}{dx}(x^n) = n \cdot x^{n-1} \):

\[ f'(x) = \frac{5x^4}{20} - \frac{4x^3}{12} \]

Simplifying this gives:

\[ f'(x) = \frac{x^4}{4} - \frac{x^3}{3} \]

### Step 2: Set the first derivative to zero to find critical points
Now, we set \( f'(x) = 0 \):

\[ \frac{x^4}{4} - \frac{x^3}{3} = 0 \]

To solve this, we can factor out \( x^3 \):

\[ x^3\left(\frac{x}{4} - \frac{1}{3}\right) = 0 \]

This gives us two factors:

1. \( x^3 = 0 \) which leads to \( x = 0 \)
2. \( \frac{x}{4} - \frac{1}{3} = 0 \) which leads to \( x = \frac{4}{3} \)

Thus, the critical points are \( x = 0 \) and \( x = \frac{4}{3} \).

### Step 3: Determine the nature of the critical points using the second derivative
Next, we find the second derivative \( f''(x) \):

\[ f''(x) = \frac{d}{dx}\left(f'(x)\right) = \frac{d}{dx}\left(\frac{x^4}{4} - \frac{x^3}{3}\right) \]

Calculating this gives:

\[ f''(x) = \frac{4x^3}{4} - \frac{3x^2}{3} = x^3 - x^2 \]

### Step 4: Set the second derivative to zero to find points of inflection
Now, we set \( f''(x) = 0 \):

\[ x^3 - x^2 = 0 \]

Factoring gives:

\[ x^2(x - 1) = 0 \]

This leads to:

1. \( x^2 = 0 \) which gives \( x = 0 \)
2. \( x - 1 = 0 \) which gives \( x = 1 \)

Thus, the points of inflection are \( x = 0 \) and \( x = 1 \).

### Step 5: Analyze the critical points using the second derivative test
We will evaluate the second derivative at the critical points:

1. For \( x = 0 \):
\[ f''(0) = 0^3 - 0^2 = 0 \quad \text{(inconclusive)} \]

2. For \( x = \frac{4}{3} \):
\[ f''\left(\frac{4}{3}\right) = \left(\frac{4}{3}\right)^3 - \left(\frac{4}{3}\right)^2 = \frac{64}{27} - \frac{16}{9} = \frac{64}{27} - \frac{48}{27} = \frac{16}{27} > 0 \quad \text{(local minimum)} \]

### Conclusion
- At \( x = 0 \), we have a local maximum (since \( f''(x) \) changes from positive to negative).
- At \( x = \frac{4}{3} \), we have a local minimum (since \( f''(x) > 0 \)).
- The point of inflection occurs at \( x = 1 \) (where \( f''(x) \) changes sign).

### Summary of Results
- Local maxima: \( x = 0 \)
- Local minima: \( x = \frac{4}{3} \)
- Point of inflection: \( x = 1 \)