Math Problem Statement

Consider the function f(x)=3/4x^4-x^3-9x^2+20 (a) Find f(-2) (b) Find f'(x) The graph of the function f has a local minimum at the point where x=-2 (c) Using your answer to part (b), show that there is a second local minimum at x = 3

Solution

Let’s solve the parts step-by-step:

(a) Find f(2)f(-2)

Given the function:
f(x)=34x4x39x2+20f(x) = \frac{3}{4}x^4 - x^3 - 9x^2 + 20 Substitute x=2x = -2 into the function:

f(2)=34(2)4(2)39(2)2+20f(-2) = \frac{3}{4}(-2)^4 - (-2)^3 - 9(-2)^2 + 20

Simplify each term: (2)4=16so34(16)=12(-2)^4 = 16 \quad \text{so} \quad \frac{3}{4}(16) = 12 (2)3=8(-2)^3 = -8 (2)2=4so9(4)=36(-2)^2 = 4 \quad \text{so} \quad 9(4) = 36

Now plug everything into the expression: f(2)=12(8)36+20f(-2) = 12 - (-8) - 36 + 20 f(2)=12+836+20=4f(-2) = 12 + 8 - 36 + 20 = 4

Thus,
f(2)=4f(-2) = 4


(b) Find f(x)f'(x)

We differentiate f(x)=34x4x39x2+20f(x) = \frac{3}{4}x^4 - x^3 - 9x^2 + 20.

Apply the power rule: ddx(xn)=nxn1\frac{d}{dx}(x^n) = nx^{n-1}.

f(x)=344x33x218x+0f'(x) = \frac{3}{4} \cdot 4x^3 - 3x^2 - 18x + 0

Simplify: f(x)=3x33x218xf'(x) = 3x^3 - 3x^2 - 18x

Thus,
f(x)=3x33x218xf'(x) = 3x^3 - 3x^2 - 18x


(c) Show that there is a second local minimum at x=3x = 3

To check for local minima, we need to find the critical points by setting f(x)=0f'(x) = 0.

3x33x218x=03x^3 - 3x^2 - 18x = 0

Factor out 3x3x:

3x(x2x6)=03x(x^2 - x - 6) = 0

Solve the quadratic inside the parentheses:

x2x6=0x^2 - x - 6 = 0

Factor:

(x3)(x+2)=0(x - 3)(x + 2) = 0

So, the solutions are:

x=3andx=2x = 3 \quad \text{and} \quad x = -2

These are the critical points. Since we are given that there is a local minimum at x=2x = -2, we now check whether x=3x = 3 also corresponds to a local minimum.


Second Derivative Test

We use the second derivative to determine if the function has a local minimum at x=3x = 3.

Start by finding f(x)f''(x).
Differentiate f(x)=3x33x218xf'(x) = 3x^3 - 3x^2 - 18x.

f(x)=9x26x18f''(x) = 9x^2 - 6x - 18

Now evaluate f(x)f''(x) at x=3x = 3:

f(3)=9(3)26(3)18f''(3) = 9(3)^2 - 6(3) - 18

Simplify:

f(3)=9(9)1818=8136=45f''(3) = 9(9) - 18 - 18 = 81 - 36 = 45

Since f(3)=45>0f''(3) = 45 > 0, the function is concave up at x=3x = 3. Therefore, there is a local minimum at x=3x = 3.


Final Answer Summary

  • (a) f(2)=4f(-2) = 4
  • (b) f(x)=3x33x218xf'(x) = 3x^3 - 3x^2 - 18x
  • (c) There is a second local minimum at x=3x = 3 since f(3)=45>0f''(3) = 45 > 0, indicating concave up.

Do you have any questions or need further details?
Here are five related questions to explore:

  1. What is the value of f(3)f(3)?
  2. How would the function behave as xx \to \infty or xx \to -\infty?
  3. Can you graph the function to visualize the local minima and maxima?
  4. How do you find points of inflection for the function?
  5. What are the x-intercepts of the original function f(x)f(x)?

Tip: Always verify the sign of the second derivative at critical points to determine if it’s a local minimum or maximum.

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Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation
Critical Points
Second Derivative Test

Formulas

Power rule: d/dx(x^n) = nx^(n-1)
Second derivative test: If f''(x) > 0 at a critical point, it is a local minimum

Theorems

Second Derivative Test

Suitable Grade Level

Grades 11-12