Question

For what value(s) of x does f(x) have a local minimum? Enter a number, a list of numbers separated by commas, or NONE.

Answer 1

Answer: A differentiable function $f(x)$ has a local minimum at the point $x_0$ if two conditions are met: the value of its first derivative is equal to zero at that point and the value of its second derivative is negative at that point.

Step-by-step explanation: The procedure for finding the local minima of the function $f(x)$ is the following.

Step 1. Find the first derivative of the function $f(x)$, denoted by $f'(x)$ according to the rules of derivation.

Step 2. Find all $x$ such that $f'(x)=0.$ Denote these solutons by $x_1, x_2\ldots$.

Step 3. Find the second derivative of the function $f(x)$, denoted by $f''(x)$. Evaluate this derivative at each point found in step 2. Only If, say $f''(x_1)>0$ then $x_1$ is the local minimum and the same goes for all other values of $x$ you found in step 2.

Answer 2

For what value(s) of x does f(x) have a local minimum?

Using the example below to explain

f(x) = x2 − 6x + 5.  

Answer:

The point x on the function f(x) is a local minimum if and only if the following conditions are satisfied

1. f'(x) = 0 (at that point df(x)/dx must be equal to zero)

2. f"(x)>0 (the second derivative of the function must be greater than zero, it must be positive)

Using the example below to explain

f(x) = x2 − 6x + 5.  

Since f'(x)= 0 and f"(x) greater than 0 (positive), then we can now confirm that the function f(x) has a local minimum at x = 3

Step-by-step explanation:

The point x on the function f(x) is a local minimum if and only if the following conditions are satisfied

1. f'(x) = 0 (at that point df(x)/dx must be equal to zero)

2. f"(x)>0 (the second derivative of the function must be greater than zero, it must be positive)

For the example above:

f(x) = x2 − 6x + 5

f'(x) = 2x - 6

Condition 1:

f'(x) = 0

So,

f'(x) = 2x - 6 = 0

Solving for x

2x - 6 = 0

2x = 6

x = 3

Therefore, at x = 3, f(x) has a critical point.

We need to determine whether it is a local minimum, local maximum or saddle point.

Condition 2:

f"(x) > 0

f"(x) = f'(f'(x)) = d/dx (2x - 6) = 2

So,

f"(x) = 2 >0

Note: in some cases we would need to substitute x into f"(x) to determine the value.

Since f'(x)= 0 and f"(x) greater than 0 (positive), then we can now confirm that the function f(x) has a local minimum at x = 3