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  1. AP Calculus
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How to find intervals where f(x)f(x)f(x) is concave up/down?

  1. Find f′′(x)f''(x)f′′(x). 2. Find where f′′(x)=0f''(x) = 0f′′(x)=0 or is undefined. 3. Test intervals using test values.
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How to find intervals where f(x)f(x)f(x) is concave up/down?

  1. Find f′′(x)f''(x)f′′(x). 2. Find where f′′(x)=0f''(x) = 0f′′(x)=0 or is undefined. 3. Test intervals using test values.

How to determine if a point is an inflection point?

  1. Find f′′(x)f''(x)f′′(x). 2. Check if f′′(x)=0f''(x) = 0f′′(x)=0 at that point. 3. Verify concavity change on either side.

Steps to determine concavity of f(x)=x4−6x2f(x)=x^4 - 6x^2f(x)=x4−6x2?

  1. Find f′(x)=4x3−12xf'(x) = 4x^3 - 12xf′(x)=4x3−12x. 2. Find f′′(x)=12x2−12f''(x) = 12x^2 - 12f′′(x)=12x2−12. 3. Solve 12x2−12=012x^2 - 12 = 012x2−12=0, giving x=±1x = \pm 1x=±1. 4. Test intervals (−∞,−1)(-\infty, -1)(−∞,−1), (−1,1)(-1, 1)(−1,1), (1,∞)(1, \infty)(1,∞).

How do you use the second derivative test to find local extrema?

  1. Find critical points where f′(x)=0f'(x) = 0f′(x)=0 or is undefined. 2. Evaluate f′′(x)f''(x)f′′(x) at each critical point. 3. If f′′(x)>0f''(x) > 0f′′(x)>0, local minimum. If f′′(x)<0f''(x) < 0f′′(x)<0, local maximum. If f′′(x)=0f''(x) = 0f′′(x)=0, test is inconclusive.

How do you find the absolute maximum and minimum of a function on a closed interval?

  1. Find critical points within the interval. 2. Evaluate the function at the critical points and endpoints of the interval. 3. The largest value is the absolute maximum, and the smallest value is the absolute minimum.

How to find the first derivative of f(x)=x3+2x2−5x+1f(x) = x^3 + 2x^2 - 5x + 1f(x)=x3+2x2−5x+1?

Apply the power rule to each term: f′(x)=3x2+4x−5f'(x) = 3x^2 + 4x - 5f′(x)=3x2+4x−5.

How to find the second derivative of f(x)=3x2+4x−5f(x) = 3x^2 + 4x - 5f(x)=3x2+4x−5?

Apply the power rule to each term: f′′(x)=6x+4f''(x) = 6x + 4f′′(x)=6x+4.

How to find the possible inflection points of f(x)=6x+4f(x) = 6x + 4f(x)=6x+4?

Set f′′(x)=0f''(x) = 0f′′(x)=0 and solve for xxx: 6x+4=06x + 4 = 06x+4=0, so x=−23x = -\frac{2}{3}x=−32​.

How to check if x=−23x = -\frac{2}{3}x=−32​ is an inflection point for f(x)=x3+2x2−5x+1f(x) = x^3 + 2x^2 - 5x + 1f(x)=x3+2x2−5x+1?

Check the sign of f′′(x)f''(x)f′′(x) on either side of x=−23x = -\frac{2}{3}x=−32​. If the sign changes, it is an inflection point.

How to determine the concavity of f(x)=x3+2x2−5x+1f(x) = x^3 + 2x^2 - 5x + 1f(x)=x3+2x2−5x+1 at x=0x = 0x=0?

Evaluate f′′(0)=6(0)+4=4f''(0) = 6(0) + 4 = 4f′′(0)=6(0)+4=4. Since f′′(0)>0f''(0) > 0f′′(0)>0, the function is concave up at x=0x = 0x=0.

What does an increasing f′(x)f'(x)f′(x) graph indicate about f(x)f(x)f(x)?

f(x)f(x)f(x) is concave up.

What does a decreasing f′(x)f'(x)f′(x) graph indicate about f(x)f(x)f(x)?

f(x)f(x)f(x) is concave down.

If the graph of f′′(x)f''(x)f′′(x) is above the x-axis, what does this say about the concavity of f(x)f(x)f(x)?

f(x)f(x)f(x) is concave up.

If the graph of f′′(x)f''(x)f′′(x) is below the x-axis, what does this say about the concavity of f(x)f(x)f(x)?

f(x)f(x)f(x) is concave down.

How can you identify a possible inflection point on the graph of f′′(x)f''(x)f′′(x)?

Look for points where the graph crosses the x-axis (i.e., where f′′(x)=0f''(x) = 0f′′(x)=0).

How can you identify intervals of concave up on the graph of f′(x)f'(x)f′(x)?

Look for intervals where the slope of f′(x)f'(x)f′(x) is positive.

How can you identify intervals of concave down on the graph of f′(x)f'(x)f′(x)?

Look for intervals where the slope of f′(x)f'(x)f′(x) is negative.

What does a horizontal tangent on the graph of f′(x)f'(x)f′(x) indicate?

A possible inflection point on the graph of f(x)f(x)f(x).

If f′(x)f'(x)f′(x) is a straight line with a positive slope, what does this indicate about f(x)f(x)f(x)?

f(x)f(x)f(x) is concave up and has a constant rate of change of its slope.

If f′(x)f'(x)f′(x) is a straight line with a negative slope, what does this indicate about f(x)f(x)f(x)?

f(x)f(x)f(x) is concave down and has a constant rate of change of its slope.

Define concavity.

The direction a curve opens; concave up faces upward, concave down faces downward.

What is a point of inflection?

A point where a function changes concavity.

Define concave up in terms of the first derivative.

The slopes of tangent lines are increasing, or f′(x)f'(x)f′(x) is increasing.

Define concave down in terms of the first derivative.

The slopes of tangent lines are decreasing, or f′(x)f'(x)f′(x) is decreasing.

How is concavity related to the second derivative?

Concave up: f′′(x)>0f''(x) > 0f′′(x)>0. Concave down: f′′(x)<0f''(x) < 0f′′(x)<0.

What is a possible point of inflection?

A point where f′′(x)=0f''(x) = 0f′′(x)=0.

What must be true at a true point of inflection?

fff must change concavity and f′′(x)=0f''(x) = 0f′′(x)=0.

What does a positive second derivative indicate?

The function is concave up.

What does a negative second derivative indicate?

The function is concave down.

How to find possible inflection points?

Set the second derivative equal to zero and solve for x: f′′(x)=0f''(x)=0f′′(x)=0.