Find the maximum and minimum values of the function on the interval
The minimum value
The maximum value

Ask by Mejia Joseph. in the United States

Mar 28,2025

Question

Find the maximum and minimum values of the function $$ f(x)=x-\frac{125 x}{x+5} $$ on the interval $$ [0,23] $$ The minimum value $$ =\square $$ The maximum value $$ =\square $$

UpStudy AI · Accepted Answer

**Step 1. Compute the derivative**

The function is
$$
f(x)=x-\frac{125x}{x+5}.
$$
Differentiate term by term. The derivative of $$x$$ is $$1$$. For the second term, use the quotient rule for
$$
g(x)=\frac{125x}{x+5}.
$$
The quotient rule gives
$$
g'(x)=\frac{125(x+5)-125x}{(x+5)^2}=\frac{625}{(x+5)^2}.
$$
Thus, the derivative of $$f(x)$$ is
$$
f'(x)=1-\frac{625}{(x+5)^2}.
$$

**Step 2. Find the critical point**

Set the derivative equal to zero:
$$
1-\frac{625}{(x+5)^2}=0.
$$
Solve for $$x$$:
$$
\frac{625}{(x+5)^2}=1\quad\Longrightarrow\quad (x+5)^2=625.
$$
Taking the positive square root (since $$x+5>0$$ in the interval $$[0,23]$$):
$$
x+5=25\quad\Longrightarrow\quad x=20.
$$
Thus, the critical point on $$[0,23]$$ is $$x=20$$.

**Step 3. Evaluate the function at the endpoints and at the critical point**

- At $$x=0$$:
  $$
  f(0)=0-\frac{125\cdot 0}{0+5}=0.
  $$
- At $$x=20$$:
  $$
  f(20)=20-\frac{125\cdot 20}{20+5}=20-\frac{2500}{25}=20-100=-80.
  $$
- At $$x=23$$:
  $$
  f(23)=23-\frac{125\cdot 23}{23+5}=23-\frac{2875}{28}.
  $$
  Calculating:
  $$
  \frac{2875}{28}\approx102.68,\quad \text{so}\quad f(23)\approx23-102.68=-79.68.
  $$

**Step 4. Determine the maximum and minimum values**

The computed values are:
- $$f(0)=0$$
- $$f(20)=-80$$
- $$f(23)\approx-79.68$$

Thus, the minimum value is at $$x=20$$ where $$f(20)=-80$$, and the maximum value is at $$x=0$$ where $$f(0)=0$$.

$$
\text{The minimum value } = -80
$$
$$
\text{The maximum value } = 0
$$
The minimum value is -80 and the maximum value is 0.

Find the maximum and minimum values of the function on the interval
The minimum value
The maximum value

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