Question

the graph of (y = f(x)) is shown below. assume the domain of (f(x)) and (f(x)) are both ((-\infty, \infty)).

remember this is the graph of (y = f(x)), not the graph of (y = f(x))

based on this graph:

(y = f(x)) is increasing on the interval(s)

(y = f(x)) is decreasing on the interval(s)

therefore (f(x)) has a max at (x =) and a local min at (x =)

(y = f(x)) is concave up on the interval(s)

(y = f(x)) is concave down on the interval(s)

therefore (f(x)) has inflection point(s) at (x =)

Explanation:

Identify intervals of increase and decrease

Using the Increasing and Decreasing Intervals and Interval Notation knowledge points

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Determine local extrema

Using the Local Extrema knowledge point

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Determine concavity intervals

The function \(y = f(x)\) is concave up where its second derivative is positive, which corresponds to where the first derivative \(f'(x)\) is increasing. Looking at the graph of \(y = f'(x)\), the curve is going upwards (increasing) for \(x > -2\).
The function \(y = f(x)\) is concave down where its second derivative is negative, which corresponds to where the first derivative \(f'(x)\) is decreasing. Looking at the graph of \(y = f'(x)\), the curve is going downwards (decreasing) for \(x < -2\).

Thus, we have:

• Concave up on: \((-2, \infty)\)
• Concave down on: \((-\infty, -2)\)

Identify inflection points

An inflection point occurs where the concavity of \(f(x)\) changes, which corresponds to a local extremum (minimum or maximum) on the graph of \(f'(x)\).
Looking at the graph of \(f'(x)\), there is a local minimum at the vertex of the parabola, which is located at \(x = -2\). Since the concavity changes from concave down to concave up at this point, \(f(x)\) has an inflection point at \(x = -2\).

Answer:

Based on the graph of \(y = f'(x)\):

• \(y = f(x)\) is increasing on the interval(s) <blank>\((-\infty, -3) \cup (-1, \infty)\)</blank>
• \(y = f(x)\) is decreasing on the interval(s) <blank>\((-3, -1)\)</blank>
• Therefore \(f(x)\) has a max at \(x =\) <blank>\(-3\)</blank> and a local min at \(x =\) <blank>\(-1\)</blank>
• \(y = f(x)\) is concave up on the interval(s) <blank>\((-2, \infty)\)</blank>
• \(y = f(x)\) is concave down on the interval(s) <blank>\((-\infty, -2)\)</blank>
• Therefore \(f(x)\) has inflection point(s) at \(x =\) <blank>\(-2\)</blank>