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Chapter 4: Problem 67

Find a cubic function \(f(x)=a x^{3}+b x^{2}+c x+d\) that has a local maximum value of 3 at \(-2\) and a local minimum value of 0 at 1 .

Short Answer

Expert verified
The cubic function is \(f(x) = x^3 - 2x^2 - 4x + 5\).

Step by step solution

01

Understand Critical Points

The function has a local maximum at \(x = -2\) with the value 3 and a local minimum at \(x = 1\) with the value 0. Therefore, our function \(f(x)\) satisfies \(f(-2) = 3\) and \(f(1) = 0\). Also, at these critical points, the derivative \(f'(x)\) should be zero: \(f'(-2) = 0\) and \(f'(1) = 0\).
02

Derivative of the Function

The derivative \(f'(x)\) of the function \(f(x) = ax^3 + bx^2 + cx + d\) is \(f'(x) = 3ax^2 + 2bx + c\). At the critical points \(x = -2\) and \(x = 1\), we have \(f'(-2) = 0\) and \(f'(1) = 0\).
03

Solve for Coefficients Using Critical Point Conditions

Substitute \(x = -2\) and \(x = 1\) into \(f'(x)\):1. \(f'(-2) = 3a(-2)^2 + 2b(-2) + c = 12a - 4b + c = 0\).2. \(f'(1) = 3a(1)^2 + 2b(1) + c = 3a + 2b + c = 0\).
04

Use Function Value Conditions

Plug \(x = -2\) and \(x = 1\) into the original function \(f(x)\):1. \(f(-2) = a(-2)^3 + b(-2)^2 + c(-2) + d = -8a + 4b - 2c + d = 3\).2. \(f(1) = a(1)^3 + b(1)^2 + c(1) + d = a + b + c + d = 0\).
05

Solve System of Equations

Solve the system of 4 equations derived from these conditions:1. \(12a - 4b + c = 0\)2. \(3a + 2b + c = 0\)3. \(-8a + 4b - 2c + d = 3\)4. \(a + b + c + d = 0\).By solving these, we derive: \(a = 1\), \(b = -2\), \(c = -4\), and \(d = 5\).
06

Final Function

Construct the cubic function using the coefficients:The cubic function is \(f(x) = x^3 - 2x^2 - 4x + 5\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Local Maximum
In the realm of calculus, a local maximum is a spot on a graph where the function reaches its highest value in a certain interval. Think of it as a peak or a hilltop where, locally, no neighboring points have a higher value. For a function like a cubic function, \(f(x) = ax^3 + bx^2 + cx + d\), finding a local maximum helps us understand where the function briefly "tops out" before dipping or changing direction.
  • If at \(x = a\), the function \(f(a)\) is larger than \(f(x)\) for all nearby x-values, then \(f(x)\) has a local maximum at \(x = a\).
  • To pinpoint this, the derivative \(f'(x)\), should equal zero at \(x = a\) because the slope flattens at the peak.
In our exercise, the local maximum occurs at \(x = -2\) with the function value of 3, which is where the function transitions from increasing to decreasing.
Local Minimum
While a local maximum describes the highest point locally, a local minimum is the lowest point in its vicinity. It's where the function dips to a trough or a valley. For a cubic function, understanding this point is essential as it shows where the function is at its lowest before potentially rising again.
  • At a local minimum, say \(x = b\), the function \(f(b)\) has a value less than \(f(x)\) for nearby x-values.
  • Just like the local maximum, the derivative \(f'(x)\) equals zero at this point because the slope changes direction from negative to positive.
In the given cubic function problem, the local minimum takes place at \(x = 1\) with a function value of 0. This is the point where the graph reaches its lowest locally before rising again.
Critical Points
Critical points are special values of x where a function's derivative is zero or undefined. For a cubic function like \(f(x) = ax^3 + bx^2 + cx + d\), these are the points we closely examine to identify local maxima or minima.
  • At these points, \(f'(x)\) equals zero, signaling a change in the graph's slope direction.
  • Contrast them to inflection points, which are broader turning points, while critical points specifically hint Peaks or valleys.
In our example problem, the critical points were identified at \(x = -2\) and \(x = 1\). At these values, the derivative \(f'(x)\) was calculated to be zero, helping further identify the positions of local peaks and valleys.
Derivative
The derivative of a function gives us the slope of the function at any point. It鈥檚 essentially the function鈥檚 rate of change. For a cubic function, the derivative is crucial in finding out where the function increases or decreases.
  • For \(f(x) = ax^3 + bx^2 + cx + d\), the derivative is \(f'(x) = 3ax^2 + 2bx + c\).
  • This derivative helps find critical points, thus identifying local maximum and minimum.
In our problem, by setting the derivative \(f'(x)\) to zero for points \(x = -2\) and \(x = 1\), we found the points where the slope is zero, aiding in discovering local extrema.
System of Equations
A system of equations is a set of simultaneous equations that are solved together. In a problem involving a cubic function, we use a system of equations to determine the unknown coefficients that define the function.
  • These equations often arise from conditions such as known function values (i.e., local maxima/minima) and derivative settings.
  • Solving these systems reveals coefficients a, b, c, and d by balancing these conditions.
For the given exercise, four equations arise from the requirements: \(f(-2) = 3\), \(f(1) = 0\), and the derivative conditions \(f'(-2) = 0\) and \(f'(1) = 0\). Solving these lead us to the coefficients \((a = 1, b = -2, c = -4, d = 5)\). This set creates the cubic function that meets all conditions.

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Most popular questions from this chapter

If a metal ball with mass \(m\) is projected in water and the force of resistance is proportional to the square of the velocity, then the distance the ball travels in time t is $$s(t)=\frac{m}{c} \ln \cosh \sqrt{\frac{g c}{m t}}$$ where \(\mathrm{C}\) is a positive constant. Find where \(c\) is a positive constant. Find \(\lim _{c \rightarrow 0^{+}} s(t).\)

If $$f(x)=\left\\{\begin{array}{ll}{\sqrt{x}} & {\text { if } x \geqslant 0} \\\ {-\sqrt{-x}} & {\text { if } x<0}\end{array}\right.$$ then the root of the equation \(f(x)=0\) is \(x=0 .\) Explain why Newton's method fails to find the root no matter which initial

\(5-8\) Use Newton's method with the specified initial approxima- tion \(x_{1}\) to find \(x_{3},\) the third approximation to the root of the given equation. (Give your answer to four decimal places.) \(\frac{1}{3} x^{3}+\frac{1}{2} x^{2}+3=0, \quad x_{1}=-3\)

Produce graphs of \(f\) that reveal all the important aspects of the curve. In particular, you should use graphs of \(f^{\prime}\) and \(f "\) to estimate the intervals of increase and decrease, extreme values, intervals of concavity, and inflection points. \(f(x)=\frac{x^{2}-1}{40 x^{3}+x+1}\)

\(23-28\) Use Newton's method to find all the roots of the equation correct to eight decimal places. Start by drawing a graph to find initial approximations. \(3 \sin \left(x^{2}\right)=2 x\)
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