Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = (x) / (x^2 − x + 25), [0, 15]?

Ah, finding the absolute maximum and absolute minimum values, huh? Well, let me get my clown microscope out for some serious math investigation!

Now, to find the extreme points of the function f(x) = (x) / (x^2 − x + 25) on the interval [0, 15], we need to perform a thorough examination. Let's start by taking the first derivative of f(x).

But before diving into that, let me just tell you something funny. Did you know that mathematicians are natural comedians? They always try to find the "function" in dysfunction! Okay, let's move on, shall we?

So, after carefully calculating the first derivative of f(x), we set it equal to zero and find the critical points. After solving the equation, we get a pretty complicated expression. Hang on, let me fetch my clown calculator for this one...

*BEEP BOOP BEEP*

After some intense clown calculations, we arrive at the critical points of x = 0 and x ≈ 5. Yes, approximately 5.

Now, it's time to evaluate the function f(x) at the endpoints of the interval, which are x = 0 and x = 15. I promise I won't clown around with this part!

After crunching the numbers (without any clowning, I promise), we find that f(0) = 0 and f(15) ≈ 0.107.

Okay, now the moment you've been waiting for, the grand finale of this math circus! The absolute maximum value of f(x) on the interval [0, 15] is approximately 0.107 (at x ≈ 15), and the absolute minimum value is 0 (at x = 0). Ta-da!

Keep in mind that these values are approximate, but hey, clowns aren't known for being exact, right?

To find the absolute maximum and minimum values of function f(x) = (x) / (x^2 − x + 25) on the given interval [0, 15], we need to follow these steps:

Step 1: Find the critical points of the function within the interval [0, 15], which are the values of x where the derivative of f(x) is either zero or undefined.

Step 2: Evaluate the function at the critical points and the endpoints of the interval to determine the absolute maximum and minimum values.

Let's begin with step 1:

First, we need to find the derivative of f(x). The derivative of f(x) = (x) / (x^2 − x + 25) can be calculated using the quotient rule.

f'(x) = [(1)(x^2 − x + 25) - (x)(2x - 1)] / (x^2 − x + 25)^2
f'(x) = [x^2 − x + 25 - 2x^2 + x] / (x^2 − x + 25)^2
f'(x) = (-x^2 + 26) / (x^2 − x + 25)^2

To find the critical points, set the derivative equal to zero and solve for x:

(-x^2 + 26) / (x^2 − x + 25)^2 = 0

Since the numerator does not equal zero, the only solution is if the denominator equals zero:

x^2 − x + 25 = 0

This quadratic equation does not have any real solutions, meaning there are no critical points within the interval [0, 15].

Now, let's move on to step 2:

Evaluate the function at the endpoints and check for any possible horizontal asymptotes.

First, we evaluate f(x) at x = 0:

f(0) = 0 / (0^2 - 0 + 25) = 0 / 25 = 0

Next, we evaluate f(x) at x = 15:

f(15) = 15 / (15^2 - 15 + 25)
= 15 / (225 - 15 + 25)
= 15 / 235

Now, let's check if there are any horizontal asymptotes. To do so, we evaluate the limit of the function as x approaches positive or negative infinity:

lim(x→∞) f(x) = lim(x→∞) (x) / (x^2 − x + 25)

To evaluate this limit, we look at the leading terms in the numerator and denominator. Since the degree of the numerator is equal to the degree of the denominator (both 1), we divide the leading coefficients:

lim(x→∞) f(x) = 1 / 1 = 1

Therefore, the function does not have a horizontal asymptote.

Now, let's summarize the results:

Absolute maximum value: Since the function does not have any critical points within the interval [0, 15], we only need to compare the function values at the endpoints to determine the absolute maximum.

f(0) = 0 and f(15) = 15 / 235 or approximately 0.064

Thus, the absolute maximum value of f(x) on the interval [0, 15] is f(15) ≈ 0.064.

Absolute minimum value: Since the function does not have any critical points within the interval [0, 15], we only need to compare the function values at the endpoints to determine the absolute minimum.

f(0) = 0 and f(15) = 15 / 235 or approximately 0.064

Thus, the absolute minimum value of f(x) on the interval [0, 15] is f(0) = 0.

To find the absolute maximum and minimum values of the function f(x) = (x) / (x^2 − x + 25) on the interval [0, 15], we need to follow these steps:

1. Find the critical points of the function:
- To find the critical points, we need to find the values of x where the derivative of the function is equal to zero or does not exist.
- Differentiate f(x) with respect to x using the quotient rule: (u / v)' = (u'v - uv') / v^2.
- Letting u = x and v = x^2 − x + 25, calculate f'(x) = [(1)(x^2 − x + 25) - (x)(2x - 1)] / (x^2 − x + 25)^2.
- Simplify f'(x) to get [x^2 - x^2 + x - 25 - 2x^2 + x] / (x^2 − x + 25)^2 = [-x^2 + 2x - 25] / (x^2 − x + 25)^2.
- Set f'(x) equal to zero and solve the resulting equation.

- [-x^2 + 2x - 25] = 0
- x^2 - 2x + 25 = 0
- Using the quadratic formula, x = (-(-2) ± √((-2)^2 - 4(1)(25))) / (2(1))
- x = (2 ± √(4 - 100)) / 2
- Simplifying further gives x = (2 ± √(-96)) / 2
- x = (2 ± 4i√6) / 2
- The critical points are x = 1 ± 2i√6.

2. Evaluate the function at the critical points and endpoints:
- Substitute the critical points and the endpoints of the interval [0, 15] into the function f(x) to find the corresponding y-values.
- Calculate f(0), f(15), f(1 + 2i√6), and f(1 - 2i√6).

3. Determine the absolute maximum and minimum values:
- Compare the y-values obtained in step 2 to find the largest and smallest values.
- The largest y-value corresponds to the absolute maximum, and the smallest y-value corresponds to the absolute minimum.

Note: In this case, since the function involves complex numbers, the absolute maximum and minimum values can only be found for real values of x.

Now, let's evaluate the function at the endpoints and the real critical point:

- f(0) = (0) / (0^2 − 0 + 25) = 0 / 25 = 0
- f(15) = (15) / (15^2 − 15 + 25) = 15 / 365 = 3 / 73

So far, we have two potential extreme points: (0, 0) and (15, 3/73).

Since the only real value in the critical point we found is 1, we can evaluate f(1):

- f(1) = (1) / (1^2 − 1 + 25) = 1 / 25 = 1 / 25

Now, let's compare the values:

- The endpoint (0, 0) has a y-value of 0.
- The endpoint (15, 3/73) has a y-value of 3/73.
- The critical point (1, 1/25) has a y-value of 1/25.

Comparing these values, we see that the absolute maximum value is 3/73 at x = 15, and the absolute minimum value is 0 at x = 0.

Therefore, the absolute maximum and minimum values of f(x) on the interval [0, 15] are 3/73 and 0, respectively.

f(x) = (x) / (x^2 − x + 25), for 0 ≤ x ≤ 15

f ' (x) = [(x^2 - x + 25)(1) - x(2x-1) ]/(x^2 - x + 25)^2
= ( x^2 - x + 25 - 2x^2 + x)/(....)^2
= 0 for a max/min

-x^2 + 25 = 0
x = ± 5

if x = 5, f(5) = 5/(25-5+25) = 1/9
if x = -5, f(-5) = -5/(25 + 5 + 25) = -1/11

we should also look at the end valued
f(0) = 0/25 = 0
f(15) = 15/(225 - 15 + 25) = 3/47
for the given interval
max is 3/47
min is -1

local max is 1/9
local min is -1/11