Rates of Change & the Derivative — Challenge
Average speed over a shorter and shorter trip settles on the speed at one instant.
What these problems cover
These questions use 9 ideas from the course. Follow any of them for a definition, where it is taught, and what it leads to.
- Rate — A ratio of two different units, like miles per hour or dollars per pound.
- Coefficient — The number multiplying a variable — the 3 in 3n.
- Function — A rule pairing each input with exactly one output.
- Slope — How much y changes for each 1 that x increases.
- Average rate of change — f(b) - f(a)b - a over an interval.
- Tangent — tan θ = oppositeadjacent.
- Derivative — The slope of the tangent, written f'(x) or dydx.
- Difference quotient — f(x+h)-f(x)h, the slope over a gap of h.
- Quotient — The answer to a division.
Problem 1
For f(x) = x3, what is the slope of the tangent at x = -2?
- 12
- -12
- -8
- 6
- -6
Worked answer
12
By the power rule, f'(x) = 3x2. At x = -2: 3(-2)2 = 3(4). = 12. It is positive because squaring removes the sign — y = x3 is rising everywhere.
Problem 2
A ball’s height is h = 48t - 16t2 feet. What is its velocity at t = 1?
- 16 ft/s
- 32 ft/s
- 48 ft/s
- 0 ft/s
- -16 ft/s
Worked answer
16 ft/s
Velocity is the derivative of height: h'(t) = 48 - 32t. At t=1: 48 - 32 = 16. Still positive, so the ball is rising — though more slowly than the 48 ft/s it started with.
Problem 3
For that same ball, when does it reach its highest point?
- t = 1.5 s
- t = 3 s
- t = 1 s
- t = 2 s
- t = 0.5 s
Worked answer
t = 1.5 s
At the top the ball is momentarily neither rising nor falling, so the velocity is zero. 48 - 32t = 0 gives t = 1.5. It lands at t = 3, so the peak is exactly halfway — as the symmetry of a parabola requires.
Problem 4
For f(x) = x2 + bx, if f'(3) = 10, what is b?
- 4
- 1
- 10
- -4
- 7
Worked answer
4
Differentiate: f'(x) = 2x + b. Set x = 3: 6 + b = 10. So b = 4.
Problem 5
What is the derivative of f(x) = 1x?
- -1x2
- 1x2
- -2x3
- 0
- ln x
Worked answer
-1x2
Write it as x-1 so the power rule applies. Bring the power down and drop it by one: -1 · x-2. That is -1x2, negative everywhere — and indeed 1x is falling on both sides of the origin.
Problem 6
At a point where f'(x) = 0, which statement is true?
- The tangent there is horizontal
- f(x) = 0 there
- The function is zero everywhere
- The curve is a straight line
- The function is undefined there
Worked answer
The tangent there is horizontal
The derivative is the slope of the tangent. A slope of zero means the tangent is flat. It says nothing about the height of the curve — a peak at y = 100 has a horizontal tangent just as much as one at y = 0.
Problem 7
For f(x) = 3x2 - 12x + 7, at which x is the curve lowest?
- 2
- -2
- 4
- 1
- 6
Worked answer
2
At the lowest point the tangent is horizontal, so the derivative is zero. f'(x) = 6x - 12 = 0. So x = 2. The coefficient of x2 is positive, so this is a minimum rather than a maximum.
Problem 8
What is the average rate of change of f(x) = x2 from x = a to x = a+h?
- 2a + h
- 2a
- h
- a2
- 2ah
Worked answer
2a + h
The average rate is f(a+h) - f(a)h = (a+h)2 - a2h. Expanding the top: a2 + 2ah + h2 - a2 = 2ah + h2. Dividing by h gives 2a + h — and letting h shrink to zero leaves 2a, the derivative.