The function 9x+3x^-1 has one local minimum x=?? and one local maximum x=??.
f[x] = 9x + 3/x
f'[x] = 9 - 3/(x^2)
f'[x] = 0 at min/max
9 - 3/(x^2) = 0
multiply across by x^2 to get
9x^2 - 3 = 0
x^2 = 1/3
x = 1/sqrt(3) or x = -1/sqrt(3)
which i will call x1 and x2.
to find which is the min and which is the max get f''[x] and sub in the two above values for x
So if f''[x1]>0 that point is a min
if f''[x1]<0 that point is a max
repeat for x2.
to find the y coordinates sub x1 and x2 back into f[x].
ok hope that helps
To find the local minimum and local maximum of the function 9x + 3x^(-1), we need to differentiate the function and set the derivative equal to zero.
Step 1: Find the derivative of the function.
Let's differentiate the function using the power rule:
f'(x) = 9 - 3x^(-2)
Step 2: Set the derivative equal to zero and solve for x.
To find the critical points, we set the derivative equal to zero and solve for x:
9 - 3x^(-2) = 0
Simplifying the equation:
9 = 3x^(-2)
Dividing both sides by 3:
3 = x^(-2)
Taking the reciprocal:
1/3 = x^2
Taking the square root of both sides:
±√(1/3) = x
The two solutions are x = √1/3 and x = -√1/3.
Step 3: Determine the nature of the critical points.
To determine whether each critical point is a local minimum, maximum, or neither, we can examine the second derivative.
Let's find the second derivative of the function:
f''(x) = 6x^(-3)
Step 4: Evaluate the second derivative at the critical points.
Substitute the critical points into the second derivative:
For x = √1/3:
f''(√1/3) = 6(√1/3)^(-3)
= 6(√1/3)^3
= 6(1/√1/3)
= 6√1/3 = 2√1
For x = -√1/3:
f''(-√1/3) = 6(-√1/3)^(-3)
= 6(-√1/3)^3
= 6(-1/√1/3)
= -6√1/3 = -2√1
Step 5: Analyze the nature of the critical points.
Since the second derivative is positive at x = √1/3 (2√1) and negative at x = -√1/3 (-2√1), we conclude:
1. x = √1/3 (2√1) is a local minimum.
2. x = -√1/3 (-2√1) is a local maximum.
Therefore, the local minimum occurs at x = 2√1 and the local maximum occurs at x = -2√1.
To find the local minimum and local maximum of the function 9x + 3x^(-1), we need to find the critical points.
Step 1: Find the derivative of the function.
Take the derivative of 9x + 3x^(-1) with respect to x:
f'(x) = 9 - 3x^(-2)
Step 2: Find the critical points.
The critical points occur when the derivative is equal to zero or undefined. In this case, the derivative is never undefined.
Setting the derivative equal to zero and solving for x:
9 - 3x^(-2) = 0
Step 3: Solve the equation.
9 = 3x^(-2)
Divide both sides by 3:
3 = x^(-2)
Take the reciprocal of both sides:
1/3 = x^2
Take the square root of both sides (considering both positive and negative roots):
x = ±√(1/3)
So, the critical points are x = √(1/3) and x = -√(1/3).
Step 4: Determine the local minimum and local maximum.
To determine if these critical points are local minimum or maximum points, we need to analyze the behavior of the function around these points.
Using the second derivative test, we take the second derivative of the function to investigate the concavity:
f''(x) = 6x^(-3)
When x = √(1/3):
f''(√(1/3)) = 6(√(1/3))^(-3) = 6(3^(1/6))^(-3) = 6(3^(-1/2)) = 6/√3 = 2√3
When x = -√(1/3):
f''(-√(1/3)) = 6(-√(1/3))^(-3) = 6(-3^(1/6))^(-3) = 6(-3^(-1/2)) = -6/√3 = -2√3
The second derivative is positive (2√3) at x = √(1/3), indicating a local minimum. The second derivative is negative (-2√3) at x = -√(1/3), indicating a local maximum.
Therefore, the function 9x + 3x^(-1) has a local minimum at x = √(1/3) and a local maximum at x = -√(1/3).