Problem 27 - Essential Calculus Early Transcendentals

Question

\(23-36=\) Find the critical numbers of the function. $$g(t)=t^{4}+t^{3}+t^{2}+1$$

Step-by-Step Solution

Answer

The critical number is \(t = 0\).

Step 1: Find the Derivative

To find the critical numbers of the function \(g(t) = t^4 + t^3 + t^2 + 1\), start by finding its derivative. The derivative \(g'(t)\) will help us find where the function has horizontal tangents or where the rate of change is zero.

Step 2: Apply Power Rule

Apply the power rule to differentiate the function term-by-term. \[ g'(t) = \frac{d}{dt}(t^4) + \frac{d}{dt}(t^3) + \frac{d}{dt}(t^2) + \frac{d}{dt}(1) \]This gives: \[ g'(t) = 4t^3 + 3t^2 + 2t \]

Step 3: Set the Derivative to Zero

To find the critical numbers, set the derivative equal to zero and solve for \(t\): \[ 4t^3 + 3t^2 + 2t = 0 \]

Step 4: Factor Out Common Terms

Factor out the common term \(t\) from the equation: \[ t(4t^2 + 3t + 2) = 0 \]

Step 5: Solve for t

The equation \(t(4t^2 + 3t + 2) = 0\) gives solutions where either \(t = 0\) or \(4t^2 + 3t + 2 = 0\). First, consider \(t = 0\) as a critical number. Next, solve for \(4t^2 + 3t + 2 = 0\) using the quadratic formula.

Step 6: Use Quadratic Formula

Apply the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to solve \(4t^2 + 3t + 2 = 0\). Here, \(a = 4\), \(b = 3\), and \(c = 2\): \[ t = \frac{-3 \pm \sqrt{3^2 - 4 \times 4 \times 2}}{2 \times 4} \]Since the discriminant \(b^2 - 4ac < 0\) (specifically \(-23\)), there are no real solutions from this quadratic equation.

Step 7: Identify the Critical Number

Since the quadratic part does not yield real roots, the only critical number is from \(t = 0\). Thus, \(t = 0\) is the only critical number for the function \(g(t) = t^4 + t^3 + t^2 + 1\).

Key Concepts

Derivative Calculation

Understanding how to find the derivative is crucial for identifying critical numbers in calculus. The derivative of a function, often denoted as \(g'(t)\), represents the rate of change or the slope of the tangent line at any point on the curve of the function. When calculating the derivative, you're essentially looking to see how fast or slow a function is changing at any given point.
In the context of the function \(g(t) = t^4 + t^3 + t^2 + 1\), the first step in finding its critical numbers is to calculate its derivative. This is because critical numbers occur where the derivative is zero or undefined. For this function, the derivative is found as \(g'(t) = 4t^3 + 3t^2 + 2t\), which simplifies our function into a form that's easier to analyze for points where it may "flatten out"—these are the potential critical numbers.

Power Rule Differentiation

The power rule is a fundamental differentiation technique used in calculus to find the derivative of polynomial terms, which is perfect for functions like \(g(t) = t^4 + t^3 + t^2 + 1\). Using the power rule, each term of the polynomial is differentiated by multiplying the term by its exponent and then reducing the exponent by one.
Here's how it works:

• For \(t^4\): Multiply by the exponent 4, and reduce the exponent by 1, resulting in \(4t^3\).
• For \(t^3\): Follow the same process to get \(3t^2\).
• For \(t^2\): This becomes \(2t\) using the same steps.
• For any constant, like 1, the derivative is 0 because a constant doesn't change.

This method provides a quick and efficient way to compute the derivative, resulting in \(g'(t) = 4t^3 + 3t^2 + 2t\). This derivative is then used to find critical numbers by solving where it equals zero.

Solving Quadratic Equations

Solving quadratic equations becomes essential when you are left with a quadratic after setting the derivative to zero. In the function explored, after factoring out \(t\), we're left with a quadratic: \(4t^2 + 3t + 2 = 0\). Unlike linear equations, quadratics can be solved using methods like factoring, completing the square, or the quadratic formula.
Here, the quadratic formula is used. The formula is \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a\), \(b\), and \(c\) are coefficients from the quadratic equation. For this particular equation:

• \(a = 4\)
• \(b = 3\)
• \(c = 2\)

Plugging these into the formula, we find the discriminant \(b^2 - 4ac\) is negative, \(-23\), indicating no real solutions. Thus, the only critical number is \(t = 0\) as derived earlier. Recognizing when a quadratic has no real solutions is just as important as finding its roots when they exist.