Math Problem Statement

The function f(x) = 2x^3 - 27x^2 + 48x + 8 has one local minimum and one local maximum. Find the local minimum and local maximum along with their values.

Solution

We are given the function:

f(x)=2x327x2+48x+8f(x) = 2x^3 - 27x^2 + 48x + 8

We need to find the local minimum and local maximum. Let's start by finding the critical points where the derivative f(x)=0f'(x) = 0.

Step 1: Find the first derivative of f(x)f(x)

f(x)=ddx(2x327x2+48x+8)=6x254x+48f'(x) = \frac{d}{dx} \left( 2x^3 - 27x^2 + 48x + 8 \right) = 6x^2 - 54x + 48

Step 2: Solve f(x)=0f'(x) = 0 to find critical points

6x254x+48=06x^2 - 54x + 48 = 0 Dividing the equation by 6: x29x+8=0x^2 - 9x + 8 = 0 Now, factor the quadratic equation: (x8)(x1)=0(x - 8)(x - 1) = 0 So, the critical points are x=8x = 8 and x=1x = 1.

Step 3: Determine whether these points are minima or maxima

We use the second derivative test.

Find the second derivative: f(x)=ddx(6x254x+48)=12x54f''(x) = \frac{d}{dx}(6x^2 - 54x + 48) = 12x - 54

Now evaluate f(x)f''(x) at the critical points:

  • For x=1x = 1: f(1)=12(1)54=1254=42f''(1) = 12(1) - 54 = 12 - 54 = -42 Since f(1)<0f''(1) < 0, there is a local maximum at x=1x = 1.

  • For x=8x = 8: f(8)=12(8)54=9654=42f''(8) = 12(8) - 54 = 96 - 54 = 42 Since f(8)>0f''(8) > 0, there is a local minimum at x=8x = 8.

Step 4: Find the values of the function at these points

To find the actual values at the local minimum and maximum, substitute x=1x = 1 and x=8x = 8 into the original function f(x)f(x).

  • For x=1x = 1: f(1)=2(1)327(1)2+48(1)+8=227+48+8=31f(1) = 2(1)^3 - 27(1)^2 + 48(1) + 8 = 2 - 27 + 48 + 8 = 31 So, the local maximum value is f(1)=31f(1) = 31.

  • For x=8x = 8: f(8)=2(8)327(8)2+48(8)+8=2(512)27(64)+48(8)+8=10241728+384+8=312f(8) = 2(8)^3 - 27(8)^2 + 48(8) + 8 = 2(512) - 27(64) + 48(8) + 8 = 1024 - 1728 + 384 + 8 = -312 So, the local minimum value is f(8)=312f(8) = -312.

Final Answer:

  • The local minimum is at x=8x = 8 with value 312-312.
  • The local maximum is at x=1x = 1 with value 3131.

Would you like a more detailed explanation or further clarification on any steps?

Here are 5 related questions you could explore:

  1. What is the graphical interpretation of local maxima and minima?
  2. How do the critical points relate to the shape of the graph of the function?
  3. What happens if both the first and second derivatives at a critical point are zero?
  4. How does one determine global maxima and minima on a closed interval?
  5. How can we apply this method to functions with more complex terms?

Tip: Always remember that a second derivative greater than zero indicates concavity upwards (local minimum), while less than zero indicates concavity downwards (local maximum).

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Math Problem Analysis

Mathematical Concepts

Calculus
Critical Points
Maxima and Minima

Formulas

f'(x) = 6x^2 - 54x + 48 (First Derivative)
f''(x) = 12x - 54 (Second Derivative)
Quadratic formula: ax^2 + bx + c = 0

Theorems

First derivative test to find critical points
Second derivative test to determine concavity (local minima/maxima)

Suitable Grade Level

Grades 10-12 or early college level