Respuesta :
Answer: A differentiable function [tex]f(x)[/tex] has a local minimum at the point [tex]x_0[/tex] 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 [tex]f(x)[/tex] is the following.
Step 1. Find the first derivative of the function [tex]f(x)[/tex], denoted by [tex]f'(x)[/tex] according to the rules of derivation.
Step 2. Find all [tex]x[/tex] such that [tex]f'(x)=0.[/tex] Denote these solutons by [tex]x_1, x_2\ldots[/tex].
Step 3. Find the second derivative of the function [tex]f(x)[/tex], denoted by [tex]f''(x)[/tex]. Evaluate this derivative at each point found in step 2. Only If, say [tex]f''(x_1)>0[/tex] then [tex]x_1[/tex] is the local minimum and the same goes for all other values of [tex]x[/tex] you found in step 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
