Notes PDF
Introductory Problems
1. If f'(a) = 0, is it true that the derivative changes sign from one side of x = a to the other?
Answer
1. Not necessarily.
Solution
1. Look at f (x) = (x-a)3.This function is a cubic, increasing towards 0 at x = a: it levels off there, but then it picks up again and continues to rise. The derivative is positive everywhere except at x = a.
2. Find the sign of the slope to cos(x) at x = 0, π/4, π/2, and π
Answer
1. f'(0) = 0, f'(π/4) < 0, f'(π/2) < 0, f'(π) = 0.
Solution
1. The function cos(x) is a wave with highest point at x = 0. From there it decreases,crossing zero at x = π/2, and reaching its minimum at x = π, where it levels off again, to turn back up. The derivatives at x = 0 and x = π are zero. The two in between, at x = π/4 and at x = π/2, are both negative.
3. Find an equation of a function whose derivative is negative for x < -5, positive for -5 < x < 5, and negative for x > 5.
Answer
1. f (x) = 1/2x3-25x.
Solution
1. We want a function whose derivative is f'(x) = (x-5)(x+5). If you graph this derivative, it’s a parabola crossing the x-axis at x = -5 and x = 5, which is exactly what we wanted. Multiplying out the parentheses, we get f'(x) = x2-25. This should remind you of the function f(x) = x3– 3x in the notes; its derivative was f'(x) = 3x2 -3. To match up the constants in our case here, we can take f(x) = 1/2x3– 25x.
4. If f'(x) periodically changes sign, does it mean that it is bounded? (meaning that it never goes above and below certain fixed values, such as cos(x))
Answer
1. No.
Solution
1. Take f(x) = x cos(x). This function oscillates up and down all the time, because of the cos(x). The usual cos(x) stays between -1 and 1, but we’re multiplying it by x which means that the amplitude goes up as we move away from x = 0. For example, cos(10π) = 10π, cos(100π) = 100π, etc.
5. Sketch a graph of f (x) = x3 + 3x2.
Answer
1. Cubic, zero at x = -3 and at x = 0, turning points (-2,4) and (0,0)
Solution
1. Let us start by rewriting the function as f (x) = x2(x + 3). This is zero at x= 0 and at x= -3. The derivative is f'(x)= 3x2+ 6x= 3x(x+2). The derivative is zero at x = 0 and at x= -2 (note the difference from f(x)= x3+ 3x, whose derivative was never zero). The function is a cubic, going up from the left, which crosses the x-axis at x = -3, and turns around at x = -2. It goes down from there: it hits the x-axis AND turns around at x = 0, and goes up from there forever.