Calculus BC ยท Unit 5
Read increase, decrease, concavity, and local extrema from the graph of f'.
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The graph in front of you may be f'
Zeros of the drawn graph are critical points of \(f\). Where the drawn graph is positive, \(f\) is increasing. Where it is negative, \(f\) is decreasing.
A sign change of f' is a turn of f
If \(f'\) changes from positive to negative, \(f\) has a local maximum. Negative to positive is a local minimum.
Slope of the f' graph is the concavity of f
Where \(f'\) is increasing, \(f\) is concave up. Where \(f'\) is decreasing, \(f\) is concave down. A change in that direction is an inflection point of \(f\).
A high point of f' is not a high point of f
A peak of \(f'\) with \(f' \neq 0\) is an inflection of \(f\), not a local extremum of \(f\).
The graph is f'. Read where f increases and where it turns.
Graph of f'
\(f\) increases where \(f' > 0\).
The graph is above the axis between the zeros \(x = -1\) and \(x = 3\).
Answer: \((-1, 3)\)
Graph of f'
At \(x = 3\), \(f'\) changes from positive to negative.
Answer: \(x = 3\)
Graph of f'
At \(x = -1\), \(f'\) changes from negative to positive.
Answer: \(x = -1\)
Graph of f'
The graph passes through \((1, 2)\), so \(f'(1) = 2 > 0\).
\(f\) is increasing at \(x = 1\).
Answer: increasing
Graph of f'
The line is below the axis on \((0, 3)\).
Answer: \((0, 3)\)
Graph of f'
\(f'\) changes from negative to positive at \(x = 3\).
Local minimum at \(x = 3\).
Answer: local min at \(x = 3\)
Concavity comes from whether the f' graph is rising or falling.
Graph of f'
\(f\) is concave up where \(f'\) is increasing.
\(f'\) increases on \((-3, 1)\).
Answer: \((-3, 1)\)
Graph of f'
\(f'\) changes from increasing to decreasing at \(x = 1\).
That is an inflection point of \(f\).
\(f'(1) = 2 \neq 0\), so \(x = 1\) is not a local extremum of \(f\).
Answer: \(x = 1\)
Graph of f'
Decreasing means \(f' < 0\). That happens on \((-3, -1)\) and \((3, 5)\).
Concave down means \(f'\) is decreasing, which is the right piece \((1, 5)\).
Both conditions hold on \((3, 5)\).
Answer: \((3, 5)\)
Graph of f'
The line is below the axis on \((0, 3)\), so \(f\) is decreasing there.
The line rises, so \(f'\) is increasing and \(f\) is concave up on \((0, 4)\).
Both hold on \((0, 3)\).
Answer: \((0, 3)\)
Graph of f'
At \(x = 1\), \(f'\) changes from increasing to decreasing and \(f'(1) \neq 0\).
That is an inflection point of \(f\).
Answer: B. inflection at \(x = 1\)
\(f' > 0\) means increasing.
\(f'' < 0\) means concave down.
Answer: increasing and concave down
A full reading of one graph, plus a table of f' and f''.
Graph of f'
Local minimum of \(f\) at \(x = -1\), where \(f'\) changes from negative to positive.
Local maximum of \(f\) at \(x = 3\), where \(f'\) changes from positive to negative.
Inflection of \(f\) at \(x = 1\), where \(f'\) changes from increasing to decreasing.
Answer: local min \(x=-1\); local max \(x=3\); inflection \(x=1\)
Graph of f'
\(f'(x) > 0\) on \((-1, 3)\) and \(f'(x) < 0\) on \((3, 5)\).
By the first derivative test, \(f\) has a local maximum at \(x = 3\).
Answer: local max at \(x = 3\)
| \(x\) | \(f'(x)\) | \(f''(x)\) |
|---|---|---|
| 1 | 0 | 5 |
| 2 | 0 | \(-4\) |
| 5 | 0 | 0 |
At \(x = 1\), \(f' = 0\) and \(f'' > 0\): local minimum.
At \(x = 2\), \(f' = 0\) and \(f'' < 0\): local maximum.
At \(x = 5\), \(f'' = 0\), so the second derivative test is inconclusive.
Answer: local min at 1; local max at 2; inconclusive at 5
The derivative is positive, so \(f\) is increasing.
The second derivative is negative, so \(f\) is concave down.
Answer: increasing and concave down
Graph of f'
The graph of \(f'\) is a line of slope 1.
\(f'\) is increasing on \((0, 4)\).
Therefore \(f\) is concave up on \((0, 4)\).
Answer: concave up on \((0, 4)\)
Graph of f'
Only at \(x = -1\) does \(f'\) change from negative to positive.
Answer: A. \(-1\)
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