Question

Properties of cubics Consider the general cubic polynomial \(f(x)=x^{3}+a x^{2}+b x+c,\) where \(a, b,\) and \(c\) are real numbers. a. Prove that \(f\) has exactly one local maximum and one local minimum provided that \(a^{2}>3 b\) b. Prove that \(f\) has no extreme values if \(a^{2}<3 b\)

Short answer

Question: Prove that a cubic function \(f(x)=x^3+ax^2+bx+c\) has exactly one local maximum and one local minimum when \(a^2 > 3b\), and no extreme values when \(a^2 < 3b\). Answer: Case A: When \(a^2 > 3b\), the cubic function has exactly one local minimum and one local maximum. Case B: When \(a^2 < 3b\), the cubic function has no extreme values.

Step-by-step solution

Find the first derivative of the cubic function

To find the local maxima and minima of the function, we need to find the first derivative of the function. The first derivative of the given cubic function \(f(x)=x^3+ax^2+bx+c\) is: \(f'(x) = \dfrac{d}{dx}(x^3+ax^2+bx+c) = 3x^2+2ax+b\)

Find critical points by setting the first derivative to zero

To find where the function has local maxima and minima or inflection points, we need to find the critical points of the function. We can do this by setting the first derivative equal to zero: \(3x^2+2ax+b=0\)

Solve the quadratic equation

Now, we need to solve this quadratic equation to find the critical points \(x_1\) and \(x_2\). It is a quadratic equation of the form \(Ax^2+Bx+C=0\). To determine the roots, we can use the discriminant \(\Delta = B^2-4AC\). Here, \(A=3\), \(B=2a\) and \(C=b\). \(\Delta = (2a)^{2} - 4(3)(b) = 4a^{2} - 12b\) Now we have two cases to analyze:

Case A: Prove that there is exactly one local minimum and one local maximum if \(a^2 > 3b\)

In this case, we have \(a^2 > 3b\) or \(4a^2 > 12b\). Therefore, the discriminant \(\Delta > 0\), which means the quadratic equation will have two distinct real roots \(x_1\) and \(x_2\). To determine if these critical points are local maxima or minima, we will apply the second derivative test. Find the second derivative of the cubic function: \(f''(x) = \dfrac{d^2}{dx^2}(x^3+ax^2+bx+c) = 6x + 2a\) Now evaluate the second derivative at the critical points \(x_1\) and \(x_2\). Since \(a^2 > 3b\), we can guarantee that the second derivative will have different signs for \(x_1\) and \(x_2\). So, one of the critical points will give a positive value (and thus be a local minimum) and the other will give a negative value (a local maximum). Thus, the cubic function will have exactly one local minimum and one local maximum when \(a^2 > 3b\).

Case B: Prove that there are no extreme values if \(a^2 < 3b\)

In this case, we have \(a^2 < 3b\). Therefore, the discriminant \(\Delta < 0\), which means the quadratic equation will have no real roots. Since there are no critical points, there will be no local maxima or minima for the cubic function. So, the cubic function will have no extreme values when \(a^2 < 3b\).

Key concepts

Critical Points

Critical points are a fundamental aspect in understanding the behavior of a polynomial function. In the context of a cubic polynomial, such as the function f(x) = x^3 + ax^2 + bx + c, critical points are values of x at which the slope of the function is zero, indicating potential maximal or minimal values. To find these points, one would calculate the first derivative of the function, f'(x) = 3x^2 + 2ax + b, and set it equal to zero.

In this context, solving 3x^2 + 2ax + b = 0 is crucial. If the resulting discriminant is positive, indicating two real solutions, the function will have two distinct critical points, which potentially could be a local maximum and a local minimum. The existence of critical points is a prerequisite for having extreme values, but not all critical points are guaranteed to be points of extremum without further analysis, which is where the second derivative test comes into play. Remember, the occurrence of critical points alone does not determine whether they are maximal or minimal values, which is why the next steps of analysis are vital.

Quadratic Equation

When dealing with cubic functions like f(x) = x^3 + ax^2 + bx + c and finding its critical points, we encounter a quadratic equation after setting the first derivative to zero. A quadratic equation is generally expressed as Ax^2 + Bx + C = 0. The solutions to this equation are the critical values we are interested in.

A key aspect when solving a quadratic equation is the discriminant, denoted as Delta, which is B^2 - 4AC. The discriminant helps us determine the nature of the roots without actually solving the equation. For the given quadratic equation derived from the first derivative of our cubic function, Delta = (2a)^2 - 4(3)(b). If Delta > 0, we have two distinct real roots, indicating two potential extreme values. Hence, analyzing the discriminant is an essential step in predicting the behavior of cubic functions at their critical points.

Second Derivative Test

The second derivative test is an essential tool for determining whether a critical point is a local maximum or minimum. Once the critical points are found, by taking the second derivative of the original function—f''(x) = 6x + 2a—and evaluating it at each critical point, we can decide the nature of these points.

Applying the Second Derivative Test

At a given critical point x, if f''(x) > 0, the function is concave up at x, and we have a local minimum. If f''(x) < 0, the function is concave down, suggesting a local maximum. In our case, the condition a^2 > 3b allows us to conclude that the second derivative will have differing signs when evaluated at the critical points due to the nature of the quadratic function produced by the first derivative. Therefore, this test not only confirms the presence of extreme values but also classifies them into maxima and minima, completing our analysis of the cubic function’s behavior at critical points.