"Calculus" — Practice

Question 1

Evaluate the limit as x tends to 0 of (1 - cos x) divided by x squared.

Accepted Answer

B

Question 2

Which statement is true for f(x) = |x| at x = 0?

Accepted Answer

B

Question 3

Evaluate the integral of x cubed times cos x from minus pi to pi.

Accepted Answer

A

Question 4

For what values of a and b is f differentiable at x = 2, where f(x) = x squared for x at most 2 and f(x) = ax + b for x greater than 2?

Accepted Answer

Two conditions are required, and solving only the first leaves the answer undetermined.

Continuity at 2: the left limit is 2 squared = 4 and the right limit is 2a + b. So 2a + b = 4.

Differentiability at 2: the left derivative is 2x evaluated at 2, which is 4, and the right derivative is a. So a = 4.

Substituting into the continuity equation gives 8 + b = 4, so b = -4.

The answer is a = 4 and b = -4.

Note what happens with a = 1 and b = 2. The continuity condition is satisfied, since 2 + 2 = 4, but the derivative jumps from 4 to 1 and the graph has a corner. This is why the two conditions must both be imposed, and why a question with two unknowns always intends differentiability rather than mere continuity.

Question 5

How many values of c in the open interval from 0 to 2 satisfy the Lagrange mean value theorem for f(x) = x cubed minus 3x?

Accepted Answer

First confirm the hypotheses. A polynomial is continuous on the closed interval and differentiable on the open one, so the theorem applies and at least one such c exists.

Compute the chord slope. f(0) = 0 and f(2) = 8 - 6 = 2, so the slope is (2 - 0) divided by (2 - 0) = 1.

Set f'(c) equal to that slope: 3c squared minus 3 = 1, so 3c squared = 4 and c squared = 4/3.

This gives c equal to plus or minus 2 over the square root of 3, approximately plus or minus 1.155.

Only the positive root lies in the open interval from 0 to 2, so exactly one value of c qualifies.

The instructive step is discarding the negative root. The theorem asserts existence of a c inside the given interval, so any algebraic solution outside it is not a valid answer, and questions frequently supply an interval that admits exactly one of two roots.

Question 6

Find the absolute maximum and minimum of f(x) = x to the fourth minus 8x squared on the closed interval from minus 1 to 3.

Accepted Answer

The function is a polynomial, so it is continuous and the extreme value theorem guarantees both extrema exist on the closed interval.

Critical points: f'(x) = 4x cubed minus 16x = 4x(x squared minus 4) = 4x(x-2)(x+2). The roots are x = 0, x = 2 and x = -2.

Only x = 0 and x = 2 lie inside the interval from minus 1 to 3; x = -2 is outside and is discarded.

Evaluate at the critical points and both endpoints, which is the complete candidate list.

f(-1) = 1 - 8 = -7.
f(0) = 0.
f(2) = 16 - 32 = -16.
f(3) = 81 - 72 = 9.

The absolute maximum is 9 at x = 3, an endpoint. The absolute minimum is -16 at x = 2, an interior critical point.

Omitting the endpoints would have given a maximum of 0, which is wrong by a wide margin, and it is exactly why endpoints belong on the candidate list.

Question 7

Evaluate the limit as x tends to 0 of (e to the x minus 1 minus x) divided by (x times sin x).

Accepted Answer

Substituting gives 0/0, so the form is indeterminate. Series expansion is faster here than two applications of L'Hopital.

Numerator: e to the x = 1 + x + x squared over 2 + x cubed over 6 + ..., so e to the x minus 1 minus x = x squared over 2 + x cubed over 6 + ....

Denominator: sin x = x - x cubed over 6 + ..., so x sin x = x squared minus x to the fourth over 6 + ....

The leading term of both numerator and denominator is of order x squared, so the limit is the ratio of those leading coefficients.

Numerator leading coefficient: 1/2. Denominator leading coefficient: 1.

The limit is 1/2.

Note how little of each expansion was needed. Since the denominator is of order x squared, expanding the numerator only to x squared would have sufficed, and carrying the cubic terms was unnecessary work. Choosing the expansion order from the denominator is what keeps this method fast.

Question 8

Classify the critical points of f(x, y) = x squared plus y squared minus 4xy.

Accepted Answer

Find the critical points by setting both first partials to zero.

f_x = 2x minus 4y = 0 gives x = 2y.
f_y = 2y minus 4x = 0 gives y = 2x.

Substituting the first into the second: y = 2(2y) = 4y, so 3y = 0 and y = 0, hence x = 0.

The only critical point is the origin.

Second partials: f_xx = 2, f_yy = 2, f_xy = -4.

Discriminant: D = (2)(2) minus (-4) squared = 4 minus 16 = -12.

Since D is negative, the origin is a saddle point.

Verify directly, which makes the saddle visible. Along the line y = 0 the function is x squared, which has a minimum at the origin. Along the line y = x the function is x squared plus x squared minus 4x squared = -2x squared, which has a maximum there. The two directions disagree, and that disagreement is precisely what a saddle is.

Question 9

For which values of p does the integral of 1 over x to the p from 0 to infinity converge?

Accepted Answer

Split the integral at 1, because the two ends fail for different reasons and must be tested separately.

From 1 to infinity, the integrand has an infinite tail. The p-test gives convergence exactly when p is greater than 1, since the antiderivative x to the (1-p) over (1-p) tends to a finite limit only when the exponent 1-p is negative.

From 0 to 1, the integrand is unbounded near the origin whenever p is positive. Here the p-test gives convergence exactly when p is less than 1, since the antiderivative must stay finite as the lower limit approaches 0.

For the full integral from 0 to infinity to converge, both pieces must converge, which requires p greater than 1 and p less than 1 simultaneously.

No value of p satisfies both, so the integral diverges for every p.

The result is worth carrying because the two conditions are opposite, and a candidate who half-remembers a single p-test will produce an answer for a question that in fact has none.

Practice — "Calculus"

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