 # Suppose f” is continuous on (−∞, ∞). If f ‘(3)=0 and f ”(3)=-3. What can you say about f? 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:

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

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

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

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.

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