Question

Consider the functions \(f(x)=x^{3}-7 x^{2}-5 x+4\) and \(f^{\prime}(x)=3 x^{2}-14 x-5\). Given that \(f\) is increasing where \(f^{\prime}(x)>0\) and \(f\) is decreasing where \(f^{\prime}(x)<0,\) find where \(f\) is increasing and where \(f\) is decreasing.

Short answer

Increasing over \((-\infty, -\frac{1}{3})\) and \((5, \infty)\); decreasing over \((-\frac{1}{3}, 5)\).

Step-by-step solution

Identify the critical points

Find where the derivative function, \(f^{\text{'}}(x)=3x^{2}-14x-5\), equals zero. Solve the equation \(3x^{2}-14x-5=0\) for \(x\) to find the critical points.

Solve the quadratic equation

Use the quadratic formula \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) where \(a=3\), \(b=-14\), and \(c=-5\). Plug these values into the formula to solve for \(x\).

Compute the discriminant

Calculate the discriminant \(\Delta = b^2 - 4ac = (-14)^2 - 4(3)(-5) = 196 + 60 = 256\). This means we will have two real roots.

Find the roots

Using the quadratic formula, the roots are:\[x = \frac{-(-14)\pm\sqrt{256}}{2(3)} = \frac{14\pm 16}{6}\]Thus, the roots are \(x = 5\) and \(x = -\frac{1}{3}\).

Test intervals around critical points

Evaluate the sign of \(f^{\text{'}}(x)\) in the intervals defined by the critical points: \(x < -\frac{1}{3}\), \(-\frac{1}{3} < x < 5\), and \(x > 5\). Choose test points in each interval to determine whether \(f^{\text{'}}(x)\) is positive or negative.

Test point 1

For \(x < -\frac{1}{3}\) (choose \(x = -1\)), \(f^{\text{'}}(-1) = 3(-1)^2 - 14(-1) - 5 = 3 + 14 - 5 = 12 > 0\), hence \(f(x)\) is increasing.

Test point 2

For \(-\frac{1}{3} < x < 5\) (choose \(x = 0\)), \(f^{\text{'}}(0) = 3(0)^2 - 14(0) - 5 = -5 < 0\), hence \(f(x)\) is decreasing.

Test point 3

For \(x > 5\) (choose \(x = 6\)), \(f^{\text{'}}(6) = 3(6)^2 - 14(6) - 5 = 108 - 84 - 5 = 19 > 0\), hence \(f(x)\) is increasing.

Conclusion

Thus, \(f(x)\) is increasing on the intervals \((-\infty, -\frac{1}{3})\) and \((5, \infty)\), and is decreasing on the interval \((-\frac{1}{3}, 5)\).

Key concepts

Critical Points

The first step in analyzing the behavior of the function is finding its critical points. These points occur where the derivative of the function is zero or undefined. Since our derivative, given by \(f^{\text{'}}(x)=3x^{2}-14x-5\), is always defined for all x, we find the critical points by setting it equal to zero: \(3x^2 - 14x - 5 = 0\).
Solving this equation gives us the values of x where the function changes its increasing or decreasing behavior. We use these critical points to test intervals around them to determine the sign of the derivative.

Derivative

The concept of a derivative is central to understanding the behavior of functions. The derivative of a function, denoted \(f^{\text{'}}(x)\), represents the rate of change of the function with respect to x. In simple terms, if \(f^{\text{'}}(x) > 0\), the function \(f(x)\) is increasing, and if \(f^{\text{'}}(x) < 0\), the function is decreasing.
For the function given, \(f(x)=x^3 - 7x^2 - 5x + 4\), the derivative is \(f^{\text{'}}(x)=3x^2 - 14x - 5\). By analyzing this derivative, we can determine where the function is increasing or decreasing.

Quadratic Formula

To solve the derivative equation \(3x^2 - 14x - 5 = 0\), we use the quadratic formula. This formula is a tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\). The quadratic formula is: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\].
In our case, \(a = 3\), \(b = -14\), and \(c = -5\). Plugging these values into the formula gives us: \[x = \frac{14 \pm \sqrt{256}}{6}\]. This results in \(x = 5\) and \(x = -\frac{1}{3}\), which are our critical points.

Intervals

Once we have our critical points, we must check the intervals around them to see where the function is increasing and decreasing. The critical points divide the x-axis into intervals. For our function, the intervals are: \((-\text{infinity}, -\frac{1}{3})\), \((- \frac{1}{3}, 5)\), and \((5, +\text{infinity})\).
We select test points within each interval to determine the sign of \(f^{\text{'}}(x)\). Testing these points:

• For \(x < -\frac{1}{3}\), \(f^{\text{'}}(x) > 0\). So, \(f(x)\) is increasing.
• For \(-\frac{1}{3} < x < 5\), \(f^{\text{'}}(x) < 0\). So, \(f(x)\) is decreasing.
• For \(x > 5\), \(f^{\text{'}}(x) > 0\). So, \(f(x)\) is increasing.

This method allows us to conclude where the function is increasing or decreasing, thus giving a complete picture of the function's behavior.