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Magazine Chapter 3: Problem 26
Find the absolute maximum and minimum values of the function, if they exist, over the indicated interval. $$ f(x)=\frac{x+2}{x^{2}+3 x+3} ; \quad[-4,4] $$
Short Answer
Expert verified
The absolute maximum is 1 at x = -1; the absolute minimum is -2/7 at x = -4.
Step by step solution
01
- Find the critical points of the function
First, we need to find the derivative of the function, \[ f(x)=\frac{x+2}{x^2+3x+3} \] Use the quotient rule: \[ f'(x)=\frac{(x^2+3x+3)'(x+2)-(x+2)'(x^2+3x+3)}{(x^2+3x+3)^2} \] Calculate the derivatives: \[ (x^2+3x+3)' = 2x+3 \] and \[ (x+2)' = 1 \] Substitute these values back: \[ f'(x)=\frac{(2x+3)(x+2)-1(x^2+3x+3)}{(x^2+3x+3)^2} \] Simplify the numerator: \[ f'(x) = \frac{2x^2 + 7x + 6 - x^2 - 3x - 3}{(x^2+3x+3)^2} = \frac{x^2 + 4x + 3}{(x^2+3x+3)^2} \] Set \[ f'(x) = 0 \] to find the critical points: \[ x^2 + 4x + 3 = 0 \] Factor the quadratic equation: \[ (x+1)(x+3) = 0 \] So, the critical points are \[ x = -1 \] and \[ x = -3 \].
02
- Evaluate the function at critical points and endpoints
Now, we need to evaluate the function at the critical points and at the endpoints of the interval. Calculate \[ f(-1) \], \[ f(-3) \], \[ f(-4) \], and \[ f(4) \]. \[ f(-1) = \frac{-1+2}{(-1)^2+3(-1)+3} = \frac{1}{1-3+3}= \frac{1}{1} = 1 \] \[ f(-3) = \frac{-3+2}{(-3)^2+3(-3)+3} = \frac{-1}{9-9+3} = \frac{-1}{3} \] \[ f(-4) = \frac{-4+2}{(-4)^2+3(-4)+3} = \frac{-2}{16-12+3} = \frac{-2}{7} \] \[ f(4) = \frac{4+2}{4^2+3(4)+3} = \frac{6}{16+12+3}= \frac{6}{31} \]
03
- Determine the absolute maximum and minimum values
Compare the obtained values to find the absolute maximum and minimum: \[ f(-1) = 1 \] \[ f(-3) = -\frac{1}{3} \] \[ f(-4) = -\frac{2}{7} \] \[ f(4) = \frac{6}{31} \] The function has an absolute maximum value of 1 at \[ x = -1 \] The function has an absolute minimum value of \[ -\frac{2}{7} \] at \[ x = -4 \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
absolute maximum
The absolute maximum value of a function is the highest point over the entire interval being considered. To find the absolute maximum, follow these steps:
- Calculate the derivative of the function.
- Find the critical points by setting the derivative equal to zero.
- Evaluate the function at these critical points and at the endpoints of the interval.
absolute minimum
The absolute minimum value of a function is the lowest point over the entire interval being considered. To identify the absolute minimum:
- Calculate the derivative of the function.
- Find the critical points by setting the derivative equal to zero.
- Evaluate the function both at the critical points and the endpoints of the interval.
critical points
Critical points are points where a function's derivative is zero or undefined. These points are potential locations for local and absolute extrema (maximum or minimum values). To find critical points:
- Compute the derivative of the function \( f(x) \).
- Set the derivative \( f'(x) \) equal to zero and solve for \( x \).
derivative
The derivative of a function measures how the function's output value changes as its input value changes. It's a core concept for finding extrema. To compute the derivative using the quotient rule (used when you have a ratio of two functions \( f = \frac{u}{v} \)):
- Identify the numerator \( u(x) = x+2 \) and the denominator \( v(x) = x^2+3x+3 \).
- Compute the derivatives \( u'(x) = 1 \) and \( v'(x) = 2x+3 \).
- Apply the quotient rule: \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \).
interval evaluation
Interval evaluation involves assessing the function at critical points and endpoints to find the absolute extrema within a specific interval. Steps include:
- Identify your interval, e.g., \( [-4, 4] \).
- Find critical points inside the interval by solving \( f'(x) = 0 \).
- Evaluate the function at these critical points and at the endpoints of the interval.
- \( f(-1) = 1 \)
- \( f(-3) = -\frac{1}{3} \)
- \( f(-4) = -\frac{2}{7} \)
- \( f(4) = \frac{6}{31} \)
