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MAT 250 Exam 2 Solutions
- For this problem, use the following function.
f (x) = x^2 − 16 x^2 − 9 x + 20
a) Find all of the discontinuities of f (x). Classify them as essential discontinuities or removable discontinuities. Solution :f (x) = ((xx−−4)(4)(xx+4)−5) This and the graph of f (x) demonstrate that there is a vertical asymptote or
essential discontinuity at x = 5 and there is a puncture or a removable discontinuity at x = 4.
b) Define f (x) so that it will be continuous at x = 4. Solution :
Let f (4) = (^) xlim→ 4 f (x) = lim x→ 4 (x + 4)(x − 4) (x − 4)(x − 5) = lim x→ 4 x + 4 x − 5
- A ball is through straight up with an initial velocity of 76 ft/sec from over the edge of a 150 foot cliff. Its height at time t is given by s(t) = − 16 t^2 + 76t + 150.
a) At what time is the ball’s velocity v(t) = 0? Solution : v(t) = s′(t) = − 32 t + 76 = 0 ⇒ 32 t = 76 or t = 19/.
b) What is the maximum height that the ball travels? Solution : s
8
8
- Find the third derivative, d
(^3) y dx^3 for^ y^ =^ √^1 4 x− 1.
y = (4x − 1)−^1 /^2 ⇒ y′^ = −2(4x − 1)−^3 /^2 ⇒ y′′^ = 12(4x − 1)−^5 /^2 ⇒ y′′′^ = −120(4x − 1)−^7 /^2
For the next two problems, use the functions f (x) = e−^6 x, g(x) = Sin−^1 (5x) and h(x) = ln(x^2 ).
- Find the derivatives of each of the functions.
Solution : f ′(x) = e−^6 x^ · (−6) = − 6 e−^6 x^ g′(x) = √ 1 −^1 (5x) 2 · (5) = √ 1 −^525 x 2 h′(x) = (^) x^12 · (2x^1 ) = (^2) x
- Using your answers to the previous problem, use the product or quotient rule to set up the following derivatives. Do not simplify your answers.
a)Solution :
d dx [h(x) · g(x)] = h′(x) · g(x) + h(x) · g′(x) = (^2) x · Sin−^1 (5x) + ln(x^2 ) · √ 1 −^525 x 2
b)Solution :
d dx
[
f (x) g(x)
]
= g(x)^ ·^ f^
′(x) − f (x) · g′(x) g(x)^2
Sin−^1 (5x)·(− 6 e−^6 x)−e−^6 x· √ 1 −^525 x 2 (Sin−^1 (2x))^2
- For this problem, use the parametric representation for the semicubical parabola, x = t^2 , y = t^3 (−∞ < t < ∞).
a) Without eliminating the parameter, find dy/dx when t=1. Solution :
dy dx
dy/dt dx/dt
3 t^2 2 t
3 t 2
t=
b) Without eliminating the parameter, find d^2 y/dx^2 when t = 1. Solution :
d^2 y dx^2
dy′/dy dx/dt
2 t
t=
- Find the equation of the line that is tangent to the curve y = x^3 + 6x − 5 tan(x) + 2 at the point (0, 2).
y′(x) = 3x^2 + 6 − 5 sec^2 (x) ⇒ y′(0) = 1. Therefore the equation is y − 2 = 1(x − 0) ⇒ y= x + 2
- A boat is being pulled toward a dock by a rope from the bow through a ring on the dock that is 6 ft. above the bow. The rope is hauled in at the rate of 2 ft/sec. How fast is the boat approaching the dock when 10 ft of rope are out? Solution : When there are h = 10 feet of rope left, the boat is x =
102 − 62 = 8 ft from the dock. Differenti- ating x^2 + 6^2 = h^2 with respect to time t gives 2x dxdt = 2h dhdt ⇒ dxdt = hx · dhdt = 108 · −2 = -5/2 ft/sec.
- Use the method of implicit differentiation to find the derivative of the function x^2 − 10 y^3 = cos 2y. One-half credit will be awarded for your answer and one-half credit will be awarded for your solution.
2 x − 30 y^2 · y′^ = − sin 2y · 2 y′^ ⇒ y′(− 30 y^2 + 2 sin 2y) = − 2 x ⇒ y′^ = (^15) y (^2) −xsin 2y
- Find the absolute maximum and the absolute minimum that the function g(θ) = sin θ + cos θ takes over the region − 2 π ≤ θ ≤ π 2. For this problem, you will be awarded one-half credit for the answer and one-half credit for the reasons for your answer. Solution : g
( (^) −π 2
= sin − 2 π + cos − 2 π = −1 whereas g
( (^) π 2
= sin π 2 + cos π 2 = 1. To find any critical points, we solve g′(θ) = cos θ − sin θ = 0 ⇒ cos θ = sin θ ⇒ 1 = tan θ ⇒ θ = π 4. Thus, g
( (^) π 4
= sin π 4 +
cos π 4 =
√ 2 2 +^
√ 2 2 =^
- Therefore, g(θ) takes an absolute minimum value of −1 at θ = − 2 π and it takes
an absolute maximum value of
2 at θ = π 4.
