This question aims to find out if the given function is **continuous** and its **first derivative** is **zero** but the **second derivative** is **non-zero** — what can we conclude about the **function?**

The question is based on the concepts of the **derivatives, second derivative test, maxima,** and **minima** of the **function.** A **local maximum** is the **highest point** on the graph of the function where the **first derivative** is **zero,** and the function starts **decreasing** after that point. A **local minimum** is the **lowest point** on the function’s graph where the **first derivative** is **zero,** and the function starts to **increase** after that point.

The **second derivative** test is performed on any given function to check for **local extremas.** The **2nd derivative test** checks whether there are **local maxima** or **local minima** at a certain **point** of the given function. Let **c** is the given point on the graph of the given **function f**, and we want to check if it contains **local maxima** or **minima.** First, we take the **first derivative** of the **function f at point c.**

\[ f'(c) = 0 \]

When the **function’s first derivative** is **zero** at **point** **c**, this means that the function has a **critical point** at **c**. Then we take the **2nd derivative** and check its value at **c**, the following three situations can occur:

\[ f'(c) = 0, \hspace{0.2in} f”(c) \lt 0 \hspace{0.2in} Local\ Maximum \]

\[ f'(c) = 0, \hspace{0.2in} f”(c) \gt 0 \hspace{0.2in} Local\ Minimum \]

\[ f'(c) = 0, \hspace{0.2in} f”(c) = 0 \hspace{0.2in} Inconclusive \]

## Expert Answer

The given information about the problem is as follows:

\[ c = 3 \]

\[ f'(3) = 0 \]

\[ f”(3) = -3 \]

As the given **function** has a **first derivative equal** to **zero,** this means that there is a **critical point** at **3**. The value of the **2nd derivative** of the given function at **c=3** is **less than zero,** which means that it **has local maxima** at **c=3.**

\[ f'(3) = 0, \hspace{0.2in} f”(3) = -3 \lt 0 \hspace{0.2in} Local\ Maximum \]

## Numerical Result

The given value of the **first derivative** of the function is** 0**, and the value of the **2nd derivative** is** less than zero**. We can conclude that:

\[ f'(3) = 0, \hspace{0.2in} f”(3) = -3 \lt 0 \hspace{0.2in} Local\ Maximum \]

## Example

The **first derivative** of the **function** **f** at **c=-2** is **0**. The value of the **second derivative** at** c=-2 is 4.** What can you conclude about this?

The given information about the above problem is given as follows:

\[ c = -2 \]

\[ f'(-2) = 0 \]

\[ f”(-2) = 4 \]

Observing the **first derivative** at **c=-2,** we can conclude that the function has a **critical point** at **c**. The given value of the **second derivative** is **greater than zero,** so we can conclude that there is a **local minima** at **c=-2** on the graph of the given function.

\[ f'(-2) = 0, \hspace{0.2in} f”(-2) = 4 \gt 0 \hspace{0.2in} Local\ Minimum \]