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Prof. Haiman Math 1A—Calculus Fall, 2006
Final Examination Solutions
- Simplify x^1 /^ ln^ x.
x^1 /^ ln^ x^ = eln^ x/^ ln^ x^ = e.
- If f (x) is continuous on [0, 2], and f (0) = 1, f (1) = 2, f (2) = 0, show that f is not one-to-one.
Pick a number between 1 and 2, say 3/2. By the Intermediate Value Theorem we must have f (x) = 3/2 for some x ∈ (0, 1) and also for some x ∈ (1, 2), so f is not one-to-one.
- Find the equation of the tangent line to x^3 + y^3 = 9 at (2, 1).
Differentiate to get 3x^2 + 3y^2 y′^ = 0. At (2, 1) this gives y′^ = −4. The tangent line is therefore y = −4(x − 2) + 1 = − 4 x + 9.
- Evaluate the limit (as a number or an infinite limit):
lim x→π/ 2
1 − sin x cos^2 x
This has the “0/0” form. L’Hospital gives
lim x→π/ 2
1 − sin x cos^2 x
= lim x→π/ 2
− cos x −2 cos x sin x
= lim x→π/ 2
2 sin x
- Evaluate the limit (as a number or an infinite limit):
lim x→+∞
(1 + 2/x)x
Observe that (1 + 2/x)x^ = ex^ ln(1+2/x). Now
xlim→∞ x^ ln(1 + 2/x) = lim x→∞ ln(1 + 2/x)/(1/x).
The last expression has “0/0” form, and is equal by L’Hospital’s rule to
lim x→∞
− 2 /(x^2 (1 + 2/x)) − 1 /x^2
= lim x→∞
2 /(1 + 2/x) = 2.
Therefore, limx→+∞(1 + 2/x)x^ = e^2.
- Find c such that the line y = x + c is a slant asymptote to the curve y = x^2 /(x + 5).
The required c is given by the limit (since it exists)
c = lim x→∞
( (^) x^2 x + 5
− x
= lim x→∞
x^2 − (x^2 + 5x) x + 5
= lim x→∞
− 5 x x + 5
- Find (d/dx)^17 (ex^ + e−x).
The higher derivatives alternate between ex^ + e−x^ and ex^ − e−x. Since 17 is odd, (d/dx)^17 (ex^ + e−x) = ex^ − e−x.
- If X and Y are functions of t related by Y = eXY^ , find X′^ when Y = 1 and Y ′^ = 3.
Differentiate to get Y ′^ = eXY^ (X′Y + XY ′). When Y = 1, we have 1 = e^1 X^ , so X = 0. Substitute Y = 1, Y ′^ = 3, X = 0 into Y ′^ = eXY^ (X′Y + XY ′) to get X′^ = 3.
- Find the point on the line x + 2y = 3 closest to the origin.
It’s easiest to minimize the square of the distance from the origin, which is x^2 + y^2. Use the equation x + 2y = 3 to express this in terms of y as (3 − 2 y)^2 + y^2 = 5y^2 − 12 y + 9. Set the derivative to zero to find the minimum at 10y − 12 = 0, y = 6/5, x = 3 − 12 /5 = 3/5. So the closest point is (3/ 5 , 6 /5).
- Find all local minima and maxima of the function f (x) = x^2 e−x, and the intervals where f is increasing or decreasing.
Differentiate to get f ′(x) = x(2 − x)e−x. Therefore f (x) is decreasing on (−∞, 0] and [2, ∞), and increasing on [0, 2] with a local (and absolute) minimum at x = 0, f (0) = 0, and a local (but not absolute) maximum at x = 2, f (2) = 4e−^2.
- Show that the equation x^3 − 3 x + 3 = 0 has exactly one real root.
Set f (x) = x^3 − 3 x + 3. Then f ′(x) = 3x^2 − 3. Therefore f (x) is decreasing on [− 1 , 1] and increasing on [1, ∞), hence f (1) = 1 is a minimum on [− 1 , ∞). This shows that there is no root in [− 1 , ∞). Furthermore, f (x) is increasing on (−∞, −1], and therefore has at most one root. Since f (−1) = 5 > 0 and (for instance) f (−3) = − 15 < 0, there is a root in the interval (− 3 , −1).
- Using Newton’s method to find an approximate solution to the equation x^3 = 2, starting with first approximation x 1 = 1, find the next approximation.
We are finding a zero of f (x) = x^3 − 2. Its derivative is f ′(x) = 3x^2. The Newton step is x 2 = x 1 − f (x 1 )/f ′(x 1 ) = 1 − (−1)/3 = 4/3.
- Find f (x) such that f ′′(x) = 1 + sin x, f (0) = 0, and f ′(0) = 0.
Antidifferentiate once to get f ′(x) = x − cos x + C. Use f ′(0) = 0 to see that C = 1. Antidifferentiate again to get f (x) = x^2 /2 + x − sin x + D. Use f (0) = 0 to see that D = 0, so f (x) = x^2 /2 + x − sin x.
- Show that
0 e
−x^2 dx ≤ (1 + e− 1 / (^4) )/2.
This is most easily done using cylindrical shells. The two curves cross at (1, 2), so the volume is
2 π
0
(x + 1 − 2 x^2 )x dx = 2π
[
x^2 2
x^3 3
x^4 2
] 1
0
2 π 3
To calculate the same thing using slices, you have to do two integrals, one for y from 0 to 1 and another for y from 1 to 2.
- For the function f (x) = 1/x, find the point c in the interval (1, 3) such that f (c) is equal to the average value of f on the interval [1, 3].
The average value is fav =
1
dx x
ln 3 2
The required c satisfies f (c) = 1/c = fav, therefore c = 2/(ln 3).
