The function f(x)=4x+9x^-1 has one local minimum and one local maximum.
Question:
This function has a local minimum at x=______ with a value _______ and a local maximum at x=________ with value _______?
To find the local minimum and local maximum of the function f(x) = 4x + 9x^(-1), we need to find the first derivative and set it equal to zero. Let's do that step-by-step:
Step 1: Find the first derivative of f(x)
f'(x) = 4 - 9(x^(-2))
Step 2: Set the first derivative equal to zero and solve for x
4 - 9(x^(-2)) = 0
Step 3: Solve for x
4 = 9(x^(-2))
4/9 = x^(-2)
(4/9)^(1/2) = x^(-1)
(2/3) = x^(-1)
1/(2/3) = x
3/2 = x
Step 4: Check the second derivative to determine whether it is a local minimum or local maximum
To do this, we find the second derivative of f(x) and evaluate it at x = 3/2.
Step 4a: Find the second derivative of f(x)
f''(x) = 18(x^(-3))
Step 4b: Evaluate the second derivative at x = 3/2
f''(3/2) = 18((3/2)^(-3))
= 18((2/3)^3)
= 18((8/27))
= 16/3
Step 5: Determine the nature of the critical points
Since the second derivative at x = 3/2 is positive (16/3 > 0), it means that f(x) has a local minimum at x = 3/2.
So the local minimum of f(x) is at x = 3/2 with a value of f(3/2). To find the value, we substitute x = 3/2 into the original function f(x):
f(3/2) = 4(3/2) + 9(3/2)^(-1)
= 6 + 9(2/3)
= 6 + 6
= 12
Therefore, the local minimum is at x = 3/2 with a value of 12.
Since there is no other critical point, there is no local maximum for this function.
To find the local minimum and local maximum of a function, we need to find the critical points. Critical points are the values of x where the first derivative is either zero or undefined.
Given the function f(x) = 4x + 9x^(-1), let's find the critical points by finding the first derivative.
Step 1: Find the first derivative of f(x):
f'(x) = d/dx (4x + 9x^(-1))
To find the derivative, we can use the power rule and sum rule of derivatives.
f'(x) = 4 - 9x^(-2)
Step 2: Set the derivative equal to zero and solve for x:
4 - 9x^(-2) = 0
Step 3: Multiply through by x^2 to get rid of the negative exponent:
4x^2 - 9 = 0
Step 4: Solve for x:
4x^2 = 9
x^2 = 9/4
x = ±√(9/4)
x = ±3/2
So, the critical points are x = 3/2 and x = -3/2.
Now, let's find the values of the function at these critical points to determine if they are local minimum or maximum.
Step 5: Substitute the critical points into the original function f(x):
For x = 3/2:
f(3/2) = 4(3/2) + 9(3/2)^(-1)
f(3/2) = 6 + 9(2/3)
f(3/2) = 6 + 6
f(3/2) = 12
For x = -3/2:
f(-3/2) = 4(-3/2) + 9(-3/2)^(-1)
f(-3/2) = -6 - 9(2/3)
f(-3/2) = -6 - 6
f(-3/2) = -12
Step 6: Analyze the values to determine if they are local minimum or maximum.
Since f(3/2) = 12 and f(-3/2) = -12,
x = 3/2 corresponds to the local maximum with a value of 12, and
x = -3/2 corresponds to the local minimum with a value of -12.
Answer:
This function has a local minimum at x = -3/2 with a value of -12, and a local maximum at x = 3/2 with a value of 12.
Take the derivative of the function f(x) and set it equal to zero. Solve for x
f'(x) = 4 - 9x^-2. = 0
4 = 9/x^2
x^2 = 9/4
x = +___ or - ___
That tells you where the maxima and minima are. The second derivative tells you which it is. It is a maximum where f'(x) = 0 and f''(x) is negative.
This is basic important stuff. You should be asking how to do them, not what the answers are. Otherwise you won't learn the subject.