Increasing and Decreasing Functions

Mathematics · Calculus & Analysis · ages 17-18

Name ______________________ Date ____________

For which values of x is y = 3 - 2x decreasing?
- for x > 0 only
- for all values of x
- for x < 1.5 only
- for no values of x
For y = x² - 10x, at what value of x does the derivative change sign?
Answer: ______________
If f'(x) is negative at one particular value of x, the function is decreasing everywhere.
Circle one: True False
The curve y = x² - 8x is decreasing for x < k. What is k?
Answer: ______________
For which values of x is y = x³ - 12x increasing?
- -2 < x < 2
- x > 2 only
- for all x
- x < -2 or x > 2
On which interval is y = x² - 4x + 7 increasing?
- x < 2
- x > 7
- x > 2
- for all x
Ben differentiates y = x² - 14x, sets 2x - 14 > 0, and concludes that the curve is increasing for x < 7. Where is the slip?
- the derivative should have been 2x + 14
- 2x - 14 > 0 gives x > 7, so that is the increasing stretch
- increasing needs the derivative to be negative, not positive
- the inequality sign flips when both sides are divided by 2
For y = x³ - 27x, at what positive value of x is the derivative zero?
Answer: ______________
For which values of x is y = 2x³ + 3x² - 12x increasing?
- x < -2 or x > 1
- -2 < x < 1
- x > 1 only
- for all x
Mia differentiates y = x³ - 48x, solves 3x² - 48 > 0 to get x² > 16, and writes the answer as x > 4. Where is the slip?
- the inequality sign flips when you take the square root, so the answer is x < 4
- the derivative should have been 3x² + 48
- she should have solved 3x² - 48 < 0 instead
- x² > 16 also holds for x < -4, so the answer has two parts

Answer key

For grown-ups. Fold this page away before handing over the rest.

Increasing and Decreasing Functions W1-mt_KOrXxQvBrZ-s1

for all values of x · dy/dx = -2, which is negative for every x, so this line falls everywhere.
5 · dy/dx = 2x - 10, which is zero at x = 5. It is negative below 5 and positive above it, so 5 is where the sign turns.
False · One reading describes one point. A curve can fall in one stretch and climb in another, and most curves do.
4 · dy/dx = 2x - 8. Solving 2x - 8 < 0 gives 2x < 8, so x < 4 and the boundary is k = 4.
x < -2 or x > 2 · dy/dx = 3x² - 12 = 3(x - 2)(x + 2), which is zero at 2 and -2. An upward parabola is positive outside its roots, so the curve rises for x < -2 and for x > 2.
x > 2 · dy/dx = 2x - 4. Solving 2x - 4 > 0 gives 2x > 4, so x > 2.
2x - 14 > 0 gives x > 7, so that is the increasing stretch · His setup is right. Adding 14 gives 2x > 14, so x > 7. Dividing by a positive 2 never flips the sign, so nothing licensed his answer.
3 · dy/dx = 3x² - 27. Setting it to zero gives 3x² = 27, so x² = 9 and x is 3 or -3. The positive one is 3.
x < -2 or x > 1 · dy/dx = 6x² + 6x - 12 = 6(x + 2)(x - 1), which is zero at -2 and 1. An upward parabola is positive outside its roots.
x² > 16 also holds for x < -4, so the answer has two parts · A square bigger than 16 means the number is further than 4 from zero in either direction. Testing x = -5 gives 25 > 16, so that stretch counts too.