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Chapter 4: Problem 43

Sketch the graph of a differentiable function \(y=f(x)\) through the point \((1,1)\) if \(f^{\prime}(1)=0\) and a. \(f^{\prime}(x)>0\) for \(x<1\) and \(f^{\prime}(x)<0\) for \(x>1\) b. \(f^{\prime}(x)<0\) for \(x<1\) and \(f^{\prime}(x)>0\) for \(x>1\) c. \(f^{\prime}(x)>0\) for \(x \neq 1\) d. \(f^{\prime}(x)<0\) for \(x \neq 1\)

Short Answer

Expert verified
a. Local maximum at \(x=1\); b. Local minimum at \(x=1\); c. Plateau at \(x=1\) but increasing elsewhere; d. Plateau at \(x=1\) but decreasing elsewhere.

Step by step solution

01

Understanding the Information

We are given a function \(y=f(x)\) that is differentiable at a point \((1, 1)\) where \(f^{\prime}(1) = 0\). This indicates that the slope of the tangent line at \(x=1\) is zero, meaning there is either a local maximum, local minimum, or an inflection point at \(x=1\) depending on the surrounding derivative information.
02

Analyzing Condition (a)

For condition (a), \(f^{\prime}(x) > 0\) for \(x < 1\) means the function is increasing left of \(x=1\), and \(f^{\prime}(x) < 0\) for \(x > 1\) means the function is decreasing right of \(x=1\). This implies there is a local maximum at \(x=1\). The graph decreases after \(x=1\) and increases before \(x=1\).
03

Analyzing Condition (b)

For condition (b), \(f^{\prime}(x) < 0\) for \(x < 1\) suggests the function is decreasing left of \(x=1\), and \(f^{\prime}(x) > 0\) for \(x > 1\) suggests the function is increasing right of \(x=1\). This implies there is a local minimum at \(x=1\). The graph increases after \(x=1\) and decreases before \(x=1\).
04

Analyzing Condition (c)

For condition (c), \(f^{\prime}(x) > 0\) for \(x eq 1\) means the function is always increasing except precisely at \(x=1\). Thus the function has a stationary point at \(x=1\) but elsewhere is monotonically increasing. The graph could have a plateau or a small flat section at \(x=1\), but no minima or maxima.
05

Analyzing Condition (d)

For condition (d), \(f^{\prime}(x) < 0\) for \(x eq 1\) implies the function is always decreasing except precisely at \(x=1\). Thus, the function has a stationary point at \(x=1\) but is otherwise monotonically decreasing. The graph could have a plateau at \(x=1\), but no minima or maxima.

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Key Concepts

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

Graph Sketching
Graph sketching involves creating a visual representation of a function based on its mathematical properties. Understanding the graph of a function provides insight into its behavior across different intervals of the domain. For differentiable functions, it involves:
  • Identifying points where the slope is zero (stationary points)
  • Recognizing where the function is increasing or decreasing
  • Detecting possible local maxima, minima, or inflection points
In the given exercise, we're asked to sketch the graph of a function through a specific point and analyze its slope at various points. A clear sketch helps interpret how the function behaves as x changes, and it can illustrate concepts like continuity and differentiability at given points.
Tangent Line Slope
The slope of a tangent line to a function at a given point is determined by its derivative at that point. For a function \(y=f(x)\), a derivative like \(f'(1)=0\) tells us the tangent is horizontal at \(x=1\). This information is crucial because:
  • A zero slope indicates potential extrema like a local maximum or minimum.
  • The sign of the derivative on either side of the point tells about its increasing or decreasing nature.
In the exercise, knowing \(f'(1)=0\) indicates there might be a local extremum at \(x=1\). Moreover, examining \(f'(x)\) for nearby values of \(x\) provides further insight into the local behavior of the graph,
Local Maximum and Minimum
Local maxima and minima occur at points where the derivative changes sign, indicating a switch from increasing to decreasing or vice versa. Here's how to identify them:
  • If \(f'(x)>0\) changes to \(f'(x)<0\) at a point, it's a local maximum.
  • If \(f'(x)<0\) changes to \(f'(x)>0\), it's a local minimum.
In case of condition (a) from our exercise, the change from positive to negative derivative signifies a local maximum at \(x=1\). Whereas in condition (b), the switch in derivative from negative to positive indicates a local minimum at the same point. A thorough analysis of these changes allows us to accurately pinpoint where local extrema might exist.
Monotonicity
Monotonicity refers to a function either consistently increasing or decreasing within an interval. This concept simplifies understanding a function's overall behavior. Here's what it means in practical terms:
  • A function is monotonically increasing if \(f'(x) > 0\) over the interval.
  • A function is monotonically decreasing if \(f'(x) < 0\) over the interval.
In the given exercise, condition (c) indicates consistent increase across the domain except at \(x=1\), suggesting monotonic increase around this stationary point. Similarly, condition (d) shows a consistent decrease, indicating monotonic decrease with a plateau at \(x=1\). Identifying these monotonic trends is vital to understanding the whole graph behavior aside from specific points of interest.

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Most popular questions from this chapter

Show that for any numbers \(a\) and \(b,\) the inequality \(|\sin b-\sin a| \leq|b-a|\) is true.

Solve the initial value problems in Exercises \(67-86\). $$ \frac{d^{2} s}{d t^{2}}=\frac{3 t}{8} ;\left.\quad \frac{d s}{d t}\right|_{t=4}=3, \quad s(4)=4 $$

How we cough a. When we cough, the trachea (windpipe) contracts to increase the velocity of the air going out. This raises the questions of how much it should contract to maximize the velocity and whether it really contracts that much when we cough. Under reasonable assumptions about the elasticity of the tracheal wall and about how the air near the wall is slowed by friction, the average flow velocity \(v\) can be modeled by the equation $$ v=c\left(r_{0}-r\right) r^{2} \mathrm{cm} / \mathrm{sec}, \quad \frac{r_{0}}{2} \leq r \leq r_{0} $$ where \(r_{0}\) is the rest radius of the trachea in centimeters and \(c\) is a positive constant whose value depends in part on the length of the trachea. Show that \(v\) is greatest when \(r=(2 / 3) r_{0},\) that is, when the trachea is about 33\(\%\) contracted. The remarkable fact is that X-ray photographs confirm that the trachea contracts about this much during a cough. b. Take \(r_{0}\) to be 0.5 and \(c\) to be 1 and graph \(v\) over the interval \(0 \leq r \leq 0.5\) . Compare what you see with the claim that \(v\) isat a maximum when \(r=(2 / 3) r_{0} .\)

Solve the initial value problems in Exercises \(67-86\). $$ \frac{d r}{d \theta}=\cos \pi \theta, \quad r(0)=1 $$

Solve the initial value problems in Exercises \(67-86\). $$ \frac{d y}{d x}=9 x^{2}-4 x+5, \quad y(-1)=0 $$
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