Math / Calculus

QuestionWhere on $(-\infty, 0)$ does $f(x)=\frac{25 x^{2}+3 x+9}{x}$ take its maximum value, and what is this maximum value?

Studdy Solution

STEP 1

1. The function $f(x) = \frac{25x^2 + 3x + 9}{x}$ is defined for $x \neq 0$ .
2. We are interested in finding the maximum value of $f(x)$ on the interval $(-\infty, 0)$ .
3. To find the maximum value, we will use calculus, specifically finding the derivative and critical points.

STEP 2

1. Simplify the function if possible.
2. Find the derivative of the function.
3. Determine critical points by setting the derivative equal to zero.
4. Evaluate the function at critical points and endpoints of the interval.
5. Determine which value is the maximum.

STEP 3

Simplify the function $f(x) = \frac{25x^2 + 3x + 9}{x}$ .
$f(x) = 25x + \frac{3x}{x} + \frac{9}{x} = 25x + 3 + \frac{9}{x}$

STEP 4

Find the derivative of the simplified function $f(x) = 25x + 3 + \frac{9}{x}$ .
$f'(x) = \frac{d}{dx}(25x) + \frac{d}{dx}(3) + \frac{d}{dx}\left(\frac{9}{x}\right)$
$f'(x) = 25 + 0 - \frac{9}{x^2}$

STEP 5

Determine critical points by setting the derivative equal to zero:
$25 - \frac{9}{x^2} = 0$
$25 = \frac{9}{x^2}$
$25x^2 = 9$
$x^2 = \frac{9}{25}$
$x = \pm \frac{3}{5}$
Since we are considering the interval $(-\infty, 0)$ , we take $x = -\frac{3}{5}$ .

STEP 6

Evaluate the function at the critical point $x = -\frac{3}{5}$ .
$f\left(-\frac{3}{5}\right) = 25\left(-\frac{3}{5}\right) + 3 + \frac{9}{-\frac{3}{5}}$
$= -15 + 3 - 15$
$= -27$

STEP 7

Since $f(x)$ approaches negative infinity as $x \to 0^-$ and as $x \to -\infty$ , the maximum value on $(-\infty, 0)$ is at the critical point $x = -\frac{3}{5}$ .
The maximum value of $f(x)$ is $\boxed{-27}$ .

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