Problem 5 - Calculus with Concepts in Calculus

Question

Determine the values of \(c\) at which \(f^{\prime}\) changes from positive to negative, or from negative to positive. $$ f(t)=\frac{t^{2}-t+1}{t^{2}+t+1} $$

Step-by-Step Solution

Answer

No real values of \(c\) exist where \(f'(t)\) changes signs.

Step 1: Differentiate the Function

First, we need to find the derivative of the given function \( f(t) = \frac{t^2 - t + 1}{t^2 + t + 1} \). We will use the quotient rule for differentiation, which is: \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \). Here, \( u(t) = t^2 - t + 1 \) and \( v(t) = t^2 + t + 1 \).

Step 2: Apply the Quotient Rule

Compute the derivatives \( u'(t) = 2t - 1 \) and \( v'(t) = 2t + 1 \). Now, apply the quotient rule:\[f'(t) = \frac{(2t-1)(t^2+t+1) - (t^2-t+1)(2t+1)}{(t^2+t+1)^2}\]

Step 3: Simplify the Derivative Expression

Expand and simplify the numerator of the derivative:\[(2t-1)(t^2+t+1) = 2t^3 + 3t^2 + t - (t^2 - t + 1)(2t + 1) = 2t^3 + t^2 + 3t + 1\]After combining terms, the simplified expression becomes:\[4t^2 - 2t + 1\]

Step 4: Find Critical Points

Set the numerator of the derivative equal to zero to find critical points.\[4t^2 - 2t + 1 = 0\]Solve this quadratic equation using the quadratic formula:\[t = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \times 4 \times 1}}{2 \times 4}\]

Step 5: Solve the Quadratic Equation

Perform the calculations for the quadratic formula:\[t = \frac{2 \pm \sqrt{4 - 16}}{8} = \frac{2 \pm \sqrt{-12}}{8}\]Since the discriminant is negative, there are no real roots. Therefore, there are no critical points where the derivative changes signs.

Step 6: Conclusion of Derivative Behavior

As the derivative does not have real zeros (the critical points), \( f'(t) \) does not change signs from positive to negative or vice versa in any real domain of \( t \). The function \( f(t) \) only changes its slope as a whole without crossing zero in its derivative.

Key Concepts

Quotient Rule

The quotient rule is a method for finding the derivative of a function that is the ratio of two other functions. When you have a function in the form of \( f(t) = \frac{u(t)}{v(t)} \), finding its derivative requires careful application of the quotient rule.
The rule can be expressed as:

• \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \)

Understanding each part: - \( u(t) \) is the numerator function. - \( v(t) \) is the denominator function. - \( u'(t) \) and \( v'(t) \) are the derivatives of \( u \) and \( v \), respectively.For the function \( f(t) = \frac{t^2 - t + 1}{t^2 + t + 1} \) in our problem, you use: - \( u = t^2 - t + 1 \)- \( v = t^2 + t + 1 \)This application of the quotient rule helps in determining how the function's rate of change behaves.

Critical Points

Critical points occur where a function's derivative is zero or undefined. These points are essential for determining where a function might change its behavior.
To find critical points:

• Set the derivative equal to zero: \( f'(t) = 0 \)
• Solve for \( t \)

In the original exercise, plugging the derivative of \( f(t) \) leads to solving the quadratic equation \( 4t^2 - 2t + 1 = 0 \). For this problem, no real solutions emerged due to a negative discriminant, which implies no real critical points. Thus, the function does not change its increasing or decreasing tendency at real number points.

Quadratic Equation

A quadratic equation is a polynomial equation of the form \( ax^2 + bx + c = 0 \).
To solve a quadratic equation, you can use the quadratic formula:

• \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

In this formula:- \( a, b, \) and \( c \) are coefficients from the standard form.- The discriminant \( b^2 - 4ac \) determines the nature of the roots:

• If positive, two distinct real roots exist.
• If zero, one real root (repeated) exists.
• If negative, there are no real roots, but two complex ones.

For \( 4t^2 - 2t + 1 = 0 \), the discriminant is negative, leading to no real roots, hence no real critical points.

Function Behavior

Understanding a function's behavior involves analyzing how it changes and progresses over its domain.
When assessing function behavior using derivatives:

• A positive derivative means the function is increasing.
• A negative derivative means the function is decreasing.
• Zero derivative implies potential maximum, minimum, or plateau.

In the absence of real critical points (as in the exercise), the function \( f(t) \) does not switch between increasing and decreasing at any real number.It is vital to check derivative value or test points around suspected regions to fully understand slope behavior, even without critical points. The function may retain consistent behavior, altering gradually instead of flipping between increasing and decreasing.