Question

homework 12 (6.10)
graphical analysis
the graph of ( h(x) ) is defined on the interval (-5,7). use the graph above to answer the following questions - no justification necessary

1. **intervals where ( h(x) ) is decreasing.
2. *( x )-coordinates of the relative minima of ( h(x)
3. instantaneous rate of change of ( h(x) ) at ( x = 1
4. **all critical points of ( h(x)
5. *intervals where ( h(x) ) is concave up
6. **( x )-coordinates of the points of inflection of ( h(x)
7. **intervals where ( h(x) ) is concave down
8. **intervals where ( h(x) ) is both concave up and decreasing
9. intervals where ( h(x) ) is increasing

Explanation:

Step1: Identify decreasing intervals for $h(x)$

$h(x)$ decreases when $h'(x) < 0$. From the graph, this occurs on $(-5, 2)$.

Step2: Find relative minima of $h(x)$

Relative minima of $h(x)$ occur where $h'(x)$ switches from negative to positive, at $x=2$.

Step3: Find instantaneous rate at $x=1$

Instantaneous rate of change of $h(x)$ is $h'(1)$. From the graph, $h'(1) = -4$.

Step4: Locate critical points of $h(x)$

Critical points are where $h'(x)=0$ or undefined: $x=-2, 2, 5$.

Step5: Identify concave up intervals for $h(x)$

$h(x)$ is concave up when $h'(x)$ is increasing, on $(-2, 5) \cup (5, 7)$.

Step6: Find inflection points of $h(x)$

Inflection points occur where $h'(x)$ changes slope, at $x=-2, 5$.

Step7: Identify concave down intervals for $h(x)$

$h(x)$ is concave down when $h'(x)$ is decreasing, on $(-5, -2)$.

Step8: Find intervals concave up + decreasing

Intersection of concave up ($(-2,5)\cup(5,7)$) and decreasing ($(-5,2)$) is $(-2, 2)$.

Step9: Identify increasing intervals for $h(x)$

$h(x)$ increases when $h'(x) > 0$, on $(2, 7)$.

Answer:

1. $(-5, 2)$
2. $x=2$
3. $-4$
4. $x=-2, 2, 5$
5. $(-2, 5) \cup (5, 7)$
6. $x=-2, 5$
7. $(-5, -2)$
8. $(-2, 2)$
9. $(2, 7)$