Question

so we now have a method of determining the concavity of a non-linear function:

when the average rate of change over equal-length input-values is increasing (for all small-length intervals), the graph of the equation is concave up.

when the average rate of change over equal-length input-values is decreasing (for all small-length intervals), the graph of the equation is concave down.

using this method of determining concavity has some issues (resolved in calculus) because there is no definite way of defining “small-length intervals”. for now, let’s use intervals of 0.1. use your calculator’s table feature to evaluate functions at small values of x.

1. determine whether the graph of the function is concave up or concave down in the interval containing the given x-value. verify graphically.

a) ( f(x) = x^2 + 3x + 4, x = 0 )
b) ( f(x) = -0.1x^2 - 5, x = 1 )

1. show that the function has a point of inflection in the interval containing the x-value. verify graphically.

a) ( f(x) = x - x^3, x = 0 )
b) ( f(x) = x^3 + x + 2, x = 0 )

Explanation:

Part 4a: \( f(x) = x^2 + 3x + 4 \), \( x = 0 \)

Step 1: Recall the concavity method

We use the average rate of change over small intervals (length 0.1 here). The average rate of change (ARC) between \( x = a \) and \( x = a + h \) is \( \frac{f(a + h) - f(a)}{h} \). We'll check ARC for intervals around \( x = 0 \), e.g., \( [-0.1, 0] \), \( [0, 0.1] \), and see if ARC is increasing (concave up) or decreasing (concave down).

Step 2: Calculate ARC for \( [-0.1, 0] \)

\( h = 0.1 \), \( a = -0.1 \), \( a + h = 0 \)
\( f(-0.1) = (-0.1)^2 + 3(-0.1) + 4 = 0.01 - 0.3 + 4 = 3.71 \)
\( f(0) = 0^2 + 3(0) + 4 = 4 \)
ARC \( = \frac{4 - 3.71}{0.1} = \frac{0.29}{0.1} = 2.9 \)

Step 3: Calculate ARC for \( [0, 0.1] \)

\( a = 0 \), \( a + h = 0.1 \)
\( f(0.1) = (0.1)^2 + 3(0.1) + 4 = 0.01 + 0.3 + 4 = 4.31 \)
ARC \( = \frac{4.31 - 4}{0.1} = \frac{0.31}{0.1} = 3.1 \)

Step 4: Analyze ARC trend

The ARC increases from 2.9 (on \( [-0.1, 0] \)) to 3.1 (on \( [0, 0.1] \)). By the given method, if ARC over equal - length intervals is increasing, the graph is concave up. Graphically, \( f(x)=x^2 + 3x + 4 \) is a parabola with \( a = 1>0 \), so it opens upwards (concave up everywhere, including around \( x = 0 \)).

Step 1: Recall the concavity method

Use average rate of change over intervals of length 0.1 around \( x = 1 \), i.e., \( [0.9, 1] \) and \( [1, 1.1] \).

Step 2: Calculate ARC for \( [0.9, 1] \)

\( h = 0.1 \), \( a = 0.9 \), \( a + h = 1 \)
\( f(0.9)=-0.1(0.9)^2 - 5=-0.1\times0.81 - 5=-0.081 - 5=-5.081 \)
\( f(1)=-0.1(1)^2 - 5=-0.1 - 5=-5.1 \)
ARC \(=\frac{f(1)-f(0.9)}{0.1}=\frac{-5.1-(-5.081)}{0.1}=\frac{-0.019}{0.1}=-0.19 \)

Step 3: Calculate ARC for \( [1, 1.1] \)

\( a = 1 \), \( a + h = 1.1 \)
\( f(1.1)=-0.1(1.1)^2 - 5=-0.1\times1.21 - 5=-0.121 - 5=-5.121 \)
ARC \(=\frac{f(1.1)-f(1)}{0.1}=\frac{-5.121-(-5.1)}{0.1}=\frac{-0.021}{0.1}=-0.21 \)

Step 4: Analyze ARC trend

The ARC decreases from - 0.19 (on \( [0.9, 1] \)) to - 0.21 (on \( [1, 1.1] \)). By the given method, if ARC over equal - length intervals is decreasing, the graph is concave down. Graphically, \( f(x)=-0.1x^2 - 5 \) is a parabola with \( a=-0.1 < 0 \), so it opens downwards (concave down everywhere, including around \( x = 1 \)).

Step 1: Recall the point of inflection condition

A point of inflection occurs where the concavity changes (i.e., the average rate of change of the function changes from increasing to decreasing or vice - versa) over intervals around the point. We'll check intervals around \( x = 0 \), e.g., \( [-0.1, 0] \), \( [0, 0.1] \), and also look at the second - derivative (for verification). First, find the first and second derivatives.
First derivative: \( f^\prime(x)=1 - 3x^2 \)
Second derivative: \( f^{\prime\prime}(x)=-6x \)

Step 2: Analyze concavity around \( x = 0 \)

• For \( x<0 \) (e.g., \( x=-0.1 \)): \( f^{\prime\prime}(-0.1)=-6\times(-0.1) = 0.6>0 \), so the function is concave up on \( (-\infty, 0) \).
• For \( x > 0 \) (e.g., \( x = 0.1 \)): \( f^{\prime\prime}(0.1)=-6\times0.1=-0.6 < 0 \), so the function is concave down on \( (0,\infty) \).

Since the concavity changes at \( x = 0 \) (from concave up to concave down), \( x = 0 \) is a point of inflection.

Step 3: Using the average rate of change method (optional verification)

Calculate ARC for \( [-0.2, -0.1] \), \( [-0.1, 0] \), \( [0, 0.1] \), \( [0.1, 0.2] \)

• \( f(-0.2)=-0.2-(-0.2)^3=-0.2 + 0.008=-0.192 \)
• \( f(-0.1)=-0.1-(-0.1)^3=-0.1 + 0.001=-0.099 \)
• \( f(0)=0 - 0=0 \)
• \( f(0.1)=0.1-(0.1)^3=0.1 - 0.001 = 0.099 \)
• \( f(0.2)=0.2-(0.2)^3=0.2 - 0.008 = 0.192 \)

ARC for \( [-0.2, -0.1] \): \( \frac{f(-0.1)-f(-0.2)}{0.1}=\frac{-0.099-(-0.192)}{0.1}=\frac{0.093}{0.1}=0.93 \)
ARC for \( [-0.1, 0] \): \( \frac{f(0)-f(-0.1)}{0.1}=\frac{0-(-0.099)}{0.1}=0.99 \)
ARC for \( [0, 0.1] \): \( \frac{f(0.1)-f(0)}{0.1}=\frac{0.099 - 0}{0.1}=0.99 \)
ARC for \( [0.1, 0.2] \): \( \frac{f(0.2)-f(0.1)}{0.1}=\frac{0.192 - 0.099}{0.1}=0.93 \)

The ARC first increases (from 0.93 to 0.99) and then decreases (from 0.99 to 0.93) around \( x = 0 \), indicating a change in the rate of change of the function, hence a point of inflection.

Answer:

Concave up

Part 4b: \( f(x)=-0.1x^2 - 5 \), \( x = 1 \)