Problem 63 - Single Variable Calculus

Question

Let \(f(x)=a x^{3}+6 x^{2}+b x+4\). Determine the constants \(a\) and \(b\) such that \(f\) has a relative minimum at \(x=-1\) and a relative maximum at \(x=2\).

Step-by-Step Solution

Answer

The given function \(f(x)=ax^3 + 6x^2 + bx + 4\) cannot have a relative minimum at \(x=-1\) and a relative maximum at \(x=2\) simultaneously, as it leads to contradictory conditions on the constants \(a\) and \(b\).

Step 1: Compute the first and second derivatives of f(x)

First find the first derivative of the given polynomial function, \(f'(x)\): \(f'(x)=\frac{d}{dx}(ax^3+6x^2+bx+4)\) Using standard derivative rules, we get: \(f'(x)=3ax^2+12x+b\) Now, find the second derivative, \(f''(x)\): \(f''(x) = \frac{d^2}{dx^2}(3ax^2+12x+b)\) Using standard derivative rules, we get: \(f''(x)=6ax+12\)

Step 2: Apply the conditions for a relative minimum at \(x=-1\)

We know that for a relative minimum, \(f'(x)=0\). So, we plug in \(x=-1\) to the equation for \(f'(x)\): \(0=f'(-1)=3a(-1)^2+12(-1)+b\) Solving for \(b\), we find: \(b=-3a+12\) Also, for a relative minimum, the second derivative must be positive. So, we plug in \(x=-1\) to the equation for \(f''(x)\): \(0< f''(-1)=6a(-1)+12\) Solving the inequality, we find: \(a>2\)

Step 3: Apply the conditions for a relative maximum at \(x=2\)

We know that for a relative maximum, \(f'(x)=0\). So, plug in \(x=2\) to the equation for \(f'(x)\): \(0=f'(2)=3a(2)^2+12(2)+b\) Solving for \(b\), we find: \(b=-12a+24\) Also, for a relative maximum, the second derivative must be negative. So, plug in \(x=2\) to the equation for \(f''(x)\): \(0> f''(2)=6a(2)+12\) Solving the inequality, we find: \(a<-1\)

Step 4: Solve for the constants \(a\) and \(b\) simultaneously

Now, we have a system of equations and inequalities to solve for \(a\) and \(b\): 1) \(b=-3a+12\) 2) \(a>2\) 3) \(b=-12a+24\) 4) \(a<-1\) Since (2) and (4) are contradictory, there is no solution that satisfies both conditions simultaneously. Therefore, the given function cannot have a relative minimum at \(x=-1\) and a relative maximum at \(x=2\).

Key Concepts

Critical Points and Extremum

When studying calculus, identifying the peaks and valleys on a curve—known as relative extrema—is crucial. Critical points are where the magic happens: they're the x-values where the first derivative of the function, representing the slope of the tangent line, equals zero or does not exist. At these points, the function can potentially have a relative maximum, where the function reaches a peak, or a relative minimum, where it dips to a trough.

For the function in our original exercise, the requirement of having a relative minimum at \(x=-1\) and a relative maximum at \(x=2\) naturally leads us to examine the function's critical points. However, as we've discovered in the provided solution, the conditions we attempted to apply led to contradictory statements regarding the value of constant \(a\), indicating that such a scenario is impossible for the given polynomial.

First Derivative Test

The First Derivative Test is an essential tool for analyzing and confirming the presence of relative extrema at critical points. It involves taking the derivative of a function and assessing its sign (positive or negative) at values just before and after the critical point.

If the derivative changes from positive to negative as you pass through the critical point, it's a sign that there's a relative maximum there—the function rises, then falls. Conversely, if the derivative changes from negative to positive, you've discovered a relative minimum—the function falls, then rises. This test provides clear evidence of the function's behavior around critical points and was applied in the problem to determine the potential of \(x=-1\) and \(x=2\) to be points of relative minimum and maximum respectively.

Second Derivative Test

While the First Derivative Test gives us clues about the function’s slope, the Second Derivative Test digs into the function’s curvature to reveal more about its concavity at a critical point. Basically, a positive second derivative at a critical point means the function creates a concave up shape—like a smiling face—indicating a relative minimum. Conversely, a negative second derivative suggests a concave down shape—like a frowning face—indicating a relative maximum.

This test provides a succinct way to classify relative extrema and is particularly useful when the first derivative's sign isn't easy to determine. However, as seen in the exercise provided, conflicting requirements for the second derivative's sign at the critical points indicated that our polynomial function couldn't satisfy the given conditions for \(a\) and thus couldn't have both a relative minimum and maximum at the specified points.

Polynomial Function Differentiation

Differentiation of polynomial functions is a straightforward, yet foundational skill in calculus. By applying basic differentiation rules—like the power rule, which states that \(d/dx [x^n] = nx^{n-1}\)—you can find the slope of the function at any point, which is fundamental for identifying critical points and analyzing extrema.

In our exercise, this involved differentiating the polynomial function \(f(x)=ax^3+6x^2+bx+4\) to obtain the first and second derivatives. These derivatives were then used for applying both the First Derivative Test and the Second Derivative Test to attempt to determine values for \(a\) and \(b\) that would satisfy the given extremum conditions, which, as seen in the solution, turned out to be an impossibility.