3. The maximum value of $$ f(x)=2 x^{3}-15 x^{2}+36 x $$ on the closed interval $$ [0,4] $$ is $$ \begin{array}{llll} ext { (A) } 28 & ext { (B) } 30 & ext { (C) } 32 & ext { (D) } 48\end{array} $$

Question

3. The maximum value of $$ f(x)=2 x^{3}-15 x^{2}+36 x $$ on the closed interval $$ [0,4] $$ is $$ \begin{array}{llll}	ext { (A) } 28 & 	ext { (B) } 30 & 	ext { (C) } 32 & 	ext { (D) } 48\end{array} $$

UpStudy AI · Accepted Answer

To find the maximum value of the function $$ f(x) = 2x^3 - 15x^2 + 36x $$ on the closed interval $$[0, 4]$$, follow these steps:

1. **Find the Critical Points:**

First, compute the derivative of $$ f(x) $$ to find the critical points.

$$
   f'(x) = \frac{d}{dx}(2x^3 - 15x^2 + 36x) = 6x^2 - 30x + 36
   $$

Set the derivative equal to zero to find critical points:

$$
   6x^2 - 30x + 36 = 0
   $$

Simplify by dividing by 6:

$$
   x^2 - 5x + 6 = 0
   $$

Factor the quadratic:

$$
   (x - 2)(x - 3) = 0 \implies x = 2 \text{ or } x = 3
   $$

2. **Evaluate $$ f(x) $$ at Critical Points and Endpoints:**

- **At $$ x = 0 $$:**
     $$
     f(0) = 2(0)^3 - 15(0)^2 + 36(0) = 0
     $$

- **At $$ x = 2 $$:**
     $$
     f(2) = 2(2)^3 - 15(2)^2 + 36(2) = 16 - 60 + 72 = 28
     $$

- **At $$ x = 3 $$:**
     $$
     f(3) = 2(3)^3 - 15(3)^2 + 36(3) = 54 - 135 + 108 = 27
     $$

- **At $$ x = 4 $$:**
     $$
     f(4) = 2(4)^3 - 15(4)^2 + 36(4) = 128 - 240 + 144 = 32
     $$

3. **Determine the Maximum Value:**

Comparing the values:
   $$
   f(0) = 0, \quad f(2) = 28, \quad f(3) = 27, \quad f(4) = 32
   $$

The maximum value of $$ f(x) $$ on the interval $$[0, 4]$$ is **32**.

**Answer:** (C) 32
The maximum value of $$ f(x) = 2x^3 - 15x^2 + 36x $$ on the interval $$[0, 4]$$ is **32**.

The maximum value of on the
closed interval is

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The maximum value of on the closed interval is