Calculus Fundamentals Trivia Quiz

Answer: Rolle's Theorem

The mean value theorem states that, if a function is differentiable between the points (a,f(a)) and (b,f(b)) and continuous at the endpoints, then f'(x) must at some x be equal to the slope of the line between the two endpoints. For example, if a car goes 95 miles an hour on average between two cities, it must have been going exactly 95 mph at some point between the cities. Rolle's theorem is the mean value theorem for f(a)=f(b)=0...f' must be 0 somewhere between any two roots of a continuously differentiable function.

Answer: 1

This is where Rolle's theorem becomes really useful. We can tell the polynomial has at least one root since it goes to infinity as x does and negative infinity as x becomes very negative, so it has to hit 0 somewhere in between. If it had more than one root, then by Rolle's theorem between the two roots the derivative of the polynomial, 5x to the fourth+12, must be 0.

But 5(x to the fourth)+12 is always positive, so there can't be two roots!

Answer: 0

Please say you did NOT try to actually find the antiderivative here! Sin(xcubed)=-Sin((-x)cubed), so the area below the axis to the left of 0 is the same as the area above the axis to the right, and vice versa, so everything cancels leaving 0. On a somewhat related note, if a teacher or professor ever gives you a really really ugly integral, try answering 0. If they ask you how you got it, smile and say 'by symmetry'.

They might be impressed!

Answer: Pi over 4

Symmetry strikes again! Call the integral I. If we replace x by (Pi over 2)-x, we see that the I is the integral of Cos(x) over ((Sin(x)+Cos(x)). Adding, 2I is the integral of (Cos(x)+Sin(x)) over (Sin(x)+Cos(x))= the integral of 1, which is Pi over 2, so I has value (Pi over 4), or about 0.78. No, I don't know what the antiderivative is, but who needs to find that anyway!

Answer: 1 only

1 Is a fundamental theorem about when a function is integrable. 2 is not necessarily true. Say a function was 0 everywhere except at a single point where it was 1. As the rectangle size gets smaller and smaller, the lower sum stays at 0 while the upper sum goes to 0 because the rectangle containing 1 is smaller and smaller. 3 is also sometimes not true, for example if the function was 0 on all the rational numbers and 1 on all the irrational numbers (any rectangle contains a point where f is 0 and a point where it is 1, so the lower integral and the upper integral don't converge to each other).

Answer: Exactly n

This is the fundamental theorem of Algebra. We don't know how many roots are real, but once we include complex roots, we can count them.

Answer: Nothing...we need more information

It pays to read the fine print on your theorems! The intermediate value theorem only applies for continuous functions, so the function could have any number of roots.

Answer: False

Surprisingly, no. For example, take the functions f(x)=0 and g(x)= e to the power of (-1 over x squared)(defining g(0)=0). f(0)=g(0)=0 and both functions derivatives of all orders equal to 0! Variations on this function can be used to "glue" together two functions in a way that is infinitely differentiable

Answer: A is true, but not B

For A: If f was NOT continuous, f(x+h)-f(x) would not approach 0, so the limit defined as the derivative could not exist. Therefore f must be continuous. For B: Take a function g to be 0 at 0 and (x squared times Sine (1 over x)) elsewhere. For x not 0, g' is 2x*Sine(1 over x)+Cos(1 over x), which oscillates wildly around 0. For x equals 0, we can compute the derivative limit directly and find it is 0.

Therefore the derivative exists but is not continuous!

Answer: Fubini

Fubini's theorem says you can do this for continuous functions and in some other cases. You can't always do this for non-continuous functions though. In fact the answer you get might depend on which order you evaluate the integrals!