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Chapter 3: Problem 14

For each function: a. Make a sign diagram for the first derivative. b. Make a sign diagram for the second derivative. c. Sketch the graph by hand, showing all relative extreme points and inflection points. $$ f(x)=2 x^{4}-8 x^{3}+30 $$

Short Answer

Expert verified
The function has a local minimum at \(x = 3\) and an inflection point at \(x = 2\).

Step by step solution

01

Find the first derivative

To understand the behavior of the function, we first need to find its first derivative. The first derivative of a function gives us information about its increasing or decreasing behavior. The given function is \(f(x) = 2x^4 - 8x^3 + 30\). Applying the power rule, the first derivative is:\[f'(x) = \frac{d}{dx}(2x^4) - \frac{d}{dx}(8x^3) + \frac{d}{dx}(30) = 8x^3 - 24x^2.\]
02

Solve for critical points

Critical points occur when the derivative is zero or undefined. Here, solve for the critical points by setting \(f'(x) = 0\):\[8x^3 - 24x^2 = 0.\]Factor out the greatest common factor:\[8x^2(x - 3) = 0.\]Solving this gives \(x = 0\) or \(x = 3\). These are the critical points.
03

Create the sign diagram for the first derivative

To create a sign diagram for \(f'(x) = 8x^3 - 24x^2\), test intervals around the critical points to determine where the function is increasing or decreasing:- Choose \(x = -1\): \(f'(-1) = 8(-1)^3 - 24(-1)^2 = -8 - 24 = -32\) (negative).- Choose \(x = 1\): \(f'(1) = 8(1)^3 - 24(1)^2 = 8 - 24 = -16\) (negative).- Choose \(x = 4\): \(f'(4) = 8(4)^3 - 24(4)^2 = 512 - 384 = 128\) (positive).From the sign diagram, \(f'(x)\) is negative to the left of \(x = 3\) and positive to the right, indicating a local minimum at \(x = 3\).
04

Find the second derivative

The second derivative provides information about concavity and inflection points. Differentiate \(f'(x) = 8x^3 - 24x^2\) to find \(f''(x)\):\[f''(x) = \frac{d}{dx}(8x^3) - \frac{d}{dx}(24x^2) = 24x^2 - 48x.\]
05

Solve for points of inflection

Set \(f''(x)\) equal to zero and solve for inflection points:\[24x^2 - 48x = 0.\]Factor out \(24x\):\[24x(x - 2) = 0.\]This gives \(x = 0\) and \(x = 2\) as potential inflection points.
06

Create the sign diagram for the second derivative

Analyze the intervals between inflection point candidates to understand where the function is concave up or down:- Choose \(x = 1\): \(f''(1) = 24(1)^2 - 48(1) = 24 - 48 = -24\) (concave down).- Choose \(x = 3\): \(f''(3) = 24(3)^2 - 48(3) = 216 - 144 = 72\) (concave up).The sign graph indicates a change of concavity at \(x = 2\), confirming it as a point of inflection.
07

Sketch the function graph with important features

Using insights from steps above: - The function decreases until \(x = 3\), then starts increasing (local minimum at \(x = 3\)).- There is a point of inflection at \(x = 2\) where the function changes from concave down to concave up.- Sketch a smooth curve that reflects these behaviors, marking points at critical and inflection points.

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Key Concepts

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

First Derivative
The first derivative of a function is used to determine the function's increasing or decreasing behavior. It provides vital insights into the slope of the tangent line at any point of the function. By differentiating a given function, you can identify such attributes. For instance, in the function \(f(x) = 2x^4 - 8x^3 + 30\), you can find the first derivative by applying the power rule to each term: \[f'(x) = 8x^3 - 24x^2\].
  • The intervals where \(f'(x) > 0\) indicate that the function is increasing.
  • The intervals where \(f'(x) < 0\) indicate that the function is decreasing.
Testing values in these intervals helps create a sign diagram, which visually represents these increases and decreases. Such a diagram was created using test points like \(x = -1, 1, 4\) near critical points found by solving \(f'(x) = 0\). Knowing that \(f'(x)\) was negative before \(x = 3\) and positive after, it indicates a local minimum at \(x = 3\).
Second Derivative
The second derivative plays a crucial role in understanding the concavity of a function and identifying inflection points. It is the derivative of the first derivative, and helps reveal changes in the function's curvature. For the function given, the second derivative is\[f''(x) = 24x^2 - 48x\].
  • If \(f''(x) > 0\), the function is concave up, resembling a smile.
  • If \(f''(x) < 0\), the function is concave down, resembling a frown.
Analyzing the sign of \(f''(x)\) over different intervals provides a sign diagram, indicating where the function flips from concave up to down. Such an analysis helps in understanding the graph's bending and contributes towards identifying inflection points where this change occurs.
Critical Points
Critical points are where a function transitions in its increase or decrease, marked by the first derivative being zero or undefined. Solving \(f'(x) = 0\) helps identify these points. For the function \(f(x) = 2x^4 - 8x^3 + 30\), setting \(8x^3 - 24x^2 = 0\) gave us critical points at \(x = 0\) and \(x = 3\). These critical points are where the function might exhibit local maxima, minima, or generate horizontal tangents.
  • Analyzing the behavior of \(f'(x)\) around these points using test values provides insights about their nature (maxima/minima).
  • For example, the function decreases to \(x = 3\) (where the first derivative changes from negative to positive) and indicates a local minimum there.
Understanding critical points is vital for graphing functions and predicting behavior changes throughout the domain.
Inflection Points
Inflection points occur where a function changes its concavity, indicating a change from being concave up to concave down or vice versa. These points can be found by solving \(f''(x) = 0\). For the given function, \(24x^2 - 48x = 0\) yields potential inflection points at \(x = 0\) and \(x = 2\).
  • To confirm actual inflection points, the sign of \(f''(x)\) is checked around these values.
  • For instance, as \(f''(x)\) changes sign around \(x = 2\), this confirms \(x = 2\) as an inflection point where the function's concavity actually changes.
Identifying these points aids immensely in sketching smoother curves for functions, highlighting key turning points visually during analysis.

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Most popular questions from this chapter

In each equation, \(x\) and \(y\) are functions of \(t\) Differentiate with respect to \(t\) to find a relation between \(d x / d t\) and \(d y / d t\). $$ x^{3}-x y=y^{3} $$

Let \(y=f(u)\) be differentiable at \(u\) and \(u=g(x)\) be differentiable at \(x\). The Chain Rule (see page 141) states that \(\frac{d y}{d x}=f^{\prime}(g(x)) \cdot g^{\prime}(x)\). Identify the "hidden assumption" in the following "proof" of the Chain Rule using the differentials \(d y=f^{\prime}(u) d u\) and \(d u=g^{\prime}(x) d x\) $$ d y=f^{\prime}(u) d u=f^{\prime}(g(x)) \cdot g^{\prime}(x) d x $$ and so $$ \frac{d y}{d x}=f^{\prime}(g(x)) \cdot g^{\prime}(x) $$

In each equation, \(x\) and \(y\) are functions of \(t\) Differentiate with respect to \(t\) to find a relation between \(d x / d t\) and \(d y / d t\). $$ x^{2}+x y=y^{2} $$

In each equation, \(x\) and \(y\) are functions of \(t\) Differentiate with respect to \(t\) to find a relation between \(d x / d t\) and \(d y / d t\). $$ x y^{2}=96 $$

GENERAL: Airplane Flight Path A plane is to take off and reach a level cruising altitude of 5 miles after a horizontal distance of 100 miles, as shown in the following diagram. Find a polynomial flight path of the form \(f(x)=a x^{3}+b x^{2}+c x+d\) by following Steps i to iv to determine the constants \(a, b, c,\) and \(d\). i. Use the fact that the plane is on the ground at \(x=\) 0 [that is, \(f(0)=0]\) to determine the value of \(d\). ii. Use the fact that the path is horizontal at \(x=0\) [that is, \(\left.f^{\prime}(0)=0\right]\) to determine the value of \(c\). iii. Use the fact that at \(x=100\) the height is 5 and the path is horizontal to determine the values of \(a\) and \(b\). State the function \(f(x)\) that you have determined. iv. Use a graphing calculator to graph your function on the window [0,100] by [0,6] to verify its shape.
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