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The solutions to exam i for math105a,c calculus i, held on october 8, 2010. It includes problems on initial value problems, finding equations of tangent lines, identifying intervals where functions increase, decrease, or are concave up or down, and finding stationary points and inflection points.
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EXAM I - OCTOBER 8, 2010
Instruction: Read each question carefully. Explain ALL your work and give reasons to support your answers.
Advice: DON’T spend too much time on a single problem.
Problems Maximum Score Your Score
Total 100
1
1.(20 pts.) (i) Is y = 2x^5 a solution of the Initial Value Problem
y′^ =^10 xy with y(1) = 2?
Justify your answer.
For y = 2x^5 , it is true that y(1) = 2(1)^5 = 2. However, the D.E.
y′^ =
10 y x
does not hold because y′^ = 2(5)x^4 = 10x^4 while (^10) xy = 10(2x
(^5) ) x = 20x^4.
Thus, y = 2x^5 is NOT a solution of this IVP.
(ii) Suppose f is a function such that f (2) = 2 and f ′(2) = 1. Let h(x) = 12 f (x) + x. Find an equation of the line tangent to the graph of h at the point x = 2.
To find the tangent line to h at x = 2, we need to find the slope which is equal to h′(2). Since h(x) = 12 f (x) + x, it follows that h′(x) = 12 f ′(x) + 1 so
h′(2) =^12 f ′(2) + 1 =^12 (1) + 1 =^32.
The tangent line contains the point (2, h(2)) = (2, 3). Thus the equation of the desired tangent line is y − 3 x − 2 =
2 or^ y^ =
2 x.
3.(16 pts.) Suppose the following is the graph of the derivative f ′^ and f (0) = 2.
1
2
0
f ’
1 2 3 4
3
(i) Find all values of x (if any) at which f (x) is a local maximum.
For x to be a local maximum, f ′^ must change sign from positive to negative, i.e., the graph of f ′^ must cross the x-axis (from above the axis to below the axis). Since the graph of f ′^ stays above the x-axis, f does NOT have any local maximum.
(ii) Find all values of x (if any) at which f has an inflection point.
The function f has an inflection point at a point x if f ′′(x) = 0 and f ′′^ changes sign. Here, f ′′^ is the slope of f ′. It follows from the graph of f ′^ that f has an inflection point at x = 1 and at x = 3.
(iii) Find all values of x in the interval 0 ≤ x ≤ 4 such that f (x) is the minimum.
Since f ′^ ≥ 0 for 0 ≤ x ≤ 4 so f never decreases. This means that the minimum value of f must occur at the initial point x = 0. Thus, f attains its minimum at x = 0.
(iv) Find the limit lim x→ 0 f^ (x) x^ − 2.
Since f (0) = 2, this limit is simply
x^ lim→ 0 f^ (x) x^ − 2 = lim x→ 0 f^ (x x)^ −−^ f 0 (0)=^ f^ ′(0).
From the graph of f ′, we know that f ′(0) = 1. Thus, lim x→ 0 f^ (x)^ −^2 x
4.(15 pts.) (i) Let
f (x) = √ (^37) x + 5
x^7 − (^) x^1.
Find the exact value of f ′(1). Show your work.
Since f (x) =7x−^1 /^3 + 5x^7 /^2 − x−^1 , it follows that
f ′(x) =7 ·
x−^4 /^3 + 5 ·
x^5 /^2 − (−1)x−^2 and so
f ′(1) =7 ·
(ii) Use the limit definition of derivative to find the exact value of g′(1) for the function g(x) = 2x^2 + x.
g′(1) = lim x→ 1 g(x x)^ −−^ g 1 (1)
= lim x→ 12 x
(^2) + x − (2(1) (^2) + (1)) x − 1 = lim x→ 12 x
(^2) + x − 3 x − 1 = lim x→ 1 (2x^ + 3)( x − x 1 −^ 1) = lim x→ 1 2 x + 3 = 5.
6.(16 pts.) An apple is thrown vertically upward (under free fall conditions). Let s(t) be the position of the apple at time t, measured from the ground. Suppose the initial position of the apple is 1 meter and its initial velocity is 6.3 meters per second. Assume the gravity is a(t) = − 9 .8 meters per second per second.
(i) What is the velocity of the apple at time t?
Note that a(t) = v′(t) where v(t) is the velocity of the apple at time t. Finding the antiderivative of a(t), we have v(t) = − 9. 8 t + C 1 for some constant C 1. The initial velocity v(0) is 6. 3 so C 1 must be 6. 3 and thus
v(t) = − 9. 8 t + 6. 3.
(ii) What is the position of the apple at time t?
Note that v(t) = s′(t) where s(t) is the position of the apple at time t. Finding the antiderivative of v(t), we have s(t) = − 9. 8 t 22 + 6. 3 t + C 2 for some constant C 2. The initial position s(0) is 1 so C 2 must be 1 and thus
s(t) = − 4. 9 t^2 + 6. 3 t + 1.
(iii) When does the apple reach its maximum height? [Hint: what is the velocity at this instance?]
The apple reaches its maximum height when v(t) = 0. From (i), v(t) = 0 implies that 9. 8 t = 6. 3 or t = 149.
(iv) When does the apple hit the ground?
The apple hits the ground when the position is 0 , i.e., s(t) = 0 or 0 = − 4. 9 t^2 +
