Question

If \(f(x)=x^{3}+3 x^{2}-45 x-6\), find the value of \(f^{\prime \prime}\) at each zero of \(f^{\prime}\), that is, at each point \(c\) where \(f^{\prime}(c)=0\).

Short answer

The values are \(f''(3) = 24\) and \(f''(-5) = -24\).

Step-by-step solution

Find the First Derivative

The first step is to find the first derivative of the function \(f(x)=x^3+3x^2-45x-6\). To do this, use the power rule for derivatives.This gives:\[f'(x) = \frac{d}{dx}(x^3) + \frac{d}{dx}(3x^2) - \frac{d}{dx}(45x) - \frac{d}{dx}(6)\]Simplifying, we have:\[f'(x) = 3x^2 + 6x - 45\].

Set the First Derivative to Zero and Solve for Critical Points

To find the zeros of \(f'(x)\), set the first derivative equal to zero:\[3x^2 + 6x - 45 = 0\].This is a quadratic equation and can be solved using the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\],where \(a = 3\), \(b = 6\), \(c = -45\).Plug in the values:\[x = \frac{-6 \pm \sqrt{6^2 - 4(3)(-45)}}{2(3)}\].Simplifying the discriminant:\[x = \frac{-6 \pm \sqrt{36 + 540}}{6}\]\[x = \frac{-6 \pm \sqrt{576}}{6}\]\[x = \frac{-6 \pm 24}{6}\].This gives solutions:\[x = 3\] and \[x = -5\].

Find the Second Derivative of the Function

Next, find the second derivative of the original function. This involves differentiating \(f'(x) = 3x^2 + 6x - 45\):\[f''(x) = \frac{d}{dx}(3x^2) + \frac{d}{dx}(6x) - \frac{d}{dx}(45)\]Simplifying gives:\[f''(x) = 6x + 6\].

Evaluate the Second Derivative at the Critical Points

The critical points found earlier were \(x = 3\) and \(x = -5\). Evaluate \(f''(x)\) at these points:1. For \(x = 3\):\[f''(3) = 6(3) + 6 = 18 + 6 = 24\].2. For \(x = -5\):\[f''(-5) = 6(-5) + 6 = -30 + 6 = -24\].

Conclusion

The values of the second derivative at the points where the first derivative is zero are \(f''(3) = 24\) and \(f''(-5) = -24\).

Key concepts

Derivatives

Derivatives are fundamental in calculus. They represent the rate at which a function is changing at any given point. For a function like \( f(x) = x^3 + 3x^2 - 45x - 6 \), calculating the derivative helps us understand how the function behaves.

There are several rules to find derivatives effectively. One common rule is the power rule. According to this rule, for \( x^n \), the derivative is \( nx^{n-1} \). Applying this rule helps us derive the first derivative, \( f'(x) = 3x^2 + 6x - 45 \). Here, the derivative tells us how the function \( f(x) \) is changing with respect to \( x \).

Critical Points

Critical points occur where the derivative of a function is zero or undefined. These are key to finding where a function could have local maxima or minima. For the function \( f(x) = x^3 + 3x^2 - 45x - 6 \), we set its first derivative to zero to find its critical points: \( 3x^2 + 6x - 45 = 0 \).

The solutions \( x = 3 \) and \( x = -5 \) are our critical points. At these points, the slope of the tangent to the function is zero, indicating potential turning points. These points can suggest peaks or troughs in the function's graph.

Quadratic Formula

The quadratic formula is a powerful tool to find the roots of a quadratic equation of the form \( ax^2 + bx + c = 0 \). This formula is given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). When dealing with the equation \( 3x^2 + 6x - 45 = 0 \), we substitute \( a = 3 \), \( b = 6 \), and \( c = -45 \).

This gives us the critical points \( x = 3 \) and \( x = -5 \). The quadratic formula helps in solving for these points easily, revealing where the function's first derivative equals zero.

Second Derivative

The second derivative of a function, denoted as \( f''(x) \), provides information about the curvature of the function's graph, indicating whether it is concave up or down. For this exercise, we have \( f''(x) = 6x + 6 \).

Computing \( f''(x) \) at the critical points \( x = 3 \) and \( x = -5 \) gives the values \( f''(3) = 24 \) and \( f''(-5) = -24 \). A positive second derivative at \( x = 3 \) means the graph is concave up, suggesting a local minimum. Conversely, a negative value at \( x = -5 \) indicates concave down, suggesting a local maximum. These insights into the behavior of the function help in further analysis and plotting.