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Magazine Chapter 4: Problem 29
25-40 Find the critical numbers of the function. $$s(t)=3 t^{4}+4 t^{3}-6 t^{2}$$
Short Answer
Expert verified
The critical numbers are \( t = 0, t = \frac{-1 + \sqrt{5}}{2}, t = \frac{-1 - \sqrt{5}}{2} \).
Step by step solution
01
Find the Derivative
To find the critical numbers, we first need to find the derivative of the function. The function given is \( s(t) = 3t^{4} + 4t^{3} - 6t^{2} \). Using the power rule, the derivative \( s'(t) \) is calculated as follows:\[ s'(t) = \frac{d}{dt} (3t^{4}) + \frac{d}{dt} (4t^{3}) - \frac{d}{dt} (6t^{2}) \]This gives us:\[ s'(t) = 12t^{3} + 12t^{2} - 12t \]
02
Set the Derivative to Zero
To find the critical numbers, set the derivative \( s'(t) \) equal to zero and solve for \( t \):\[ 12t^{3} + 12t^{2} - 12t = 0 \]
03
Factor the Derivative
Factor out the greatest common factor from the derivative:\[ 12t(t^{2} + t - 1) = 0 \]Setting each factor equal to zero gives:1. \( 12t = 0 \)2. \( t^{2} + t - 1 = 0 \)
04
Solve the Equations
Solve the equations obtained from the factors:1. Solving \( 12t = 0 \) gives \( t = 0 \).2. To solve \( t^{2} + t - 1 = 0 \), use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 1, c = -1 \):\[ t = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]\[ t = \frac{-1 \pm \sqrt{5}}{2} \]Thus, the solutions are \( t = \frac{-1 + \sqrt{5}}{2} \) and \( t = \frac{-1 - \sqrt{5}}{2} \).
05
List the Critical Numbers
The critical numbers of the function \( s(t) = 3t^{4} + 4t^{3} - 6t^{2} \) are the solutions we found for \( t \):\( t = 0, t = \frac{-1 + \sqrt{5}}{2}, t = \frac{-1 - \sqrt{5}}{2} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Critical Numbers
Critical numbers are points on a graph where the slope is either zero or undefined. These points are significant because they can help you identify potential maximum, minimum, or saddle points of a function. To find these in a function, you need to:
- Calculate the derivative of the function.
- Set the derivative equal to zero.
- Solve for the variable to find the critical numbers.
Derivative
A derivative represents the rate at which a function is changing at any given point. It's like measuring how steep a hill is at different places. In mathematical terms, the derivative of a function gives its slope.
Here's how you find a derivative:
Here's how you find a derivative:
- Identify each term of the function.
- Apply the power rule to find the derivative of each term.
- Combine these results to get the final derivative.
Power Rule
The power rule is a fundamental tool in calculus for finding derivatives. It states that for a function of the form \( t^n \), the derivative is \( nt^{n-1} \). This makes taking derivatives straightforward:
- Multiply the exponent by the coefficient in front of \( t \).
- Subtract one from the exponent.
Quadratic Formula
The quadratic formula is used to find the roots of quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). The formula is:\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula will give you the values of \( t \) that satisfy the equation.
In our solution, the quadratic formula was used to solve the equation \( t^2 + t - 1 = 0 \). Here's the breakdown:
In our solution, the quadratic formula was used to solve the equation \( t^2 + t - 1 = 0 \). Here's the breakdown:
- Identify \( a = 1 \), \( b = 1 \), and \( c = -1 \).
- Substitute these values into the formula.
- Simplify to find the roots, which are the critical numbers.
