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Study Guide E
(MAT250 - 1E1)
Find the limit.
- lim x→- 6 +
(x + 1) x^ +^6 x + 6
Find the indicated limit.
- lim x→-∞
x^3 ex
Find the limit.
- lim x → - 7 +
x (^) + 7
- lim x→ 0 +
(1 (^) + csc x)
Find a simple basic function as a right-end behavior model and a simple basic function as a left-end behavior model.
y (^) = e-x^ + 2x
y = x
Answer the question.
- f(x) =
x^3 ,
- 3x, 4, 0, -^2 < x^ ≤ 0 0 ≤ x < 2 2 < x ≤ 4 x = 2
-5 -4 -3 -2 -1 1 2 3 4 5 t
10 d 8 6 4 2
(2, 0) (^) t -5 -4 -3 -2 -1 1 2 3 4 5
10 d 8 6 4 2
(2, 0)
Does lim x→ 2
f(x) (^) = f(2)?
Find the intervals on which the function is continuous.
- y (^) = 3 (x (^) + 2)2^ + 4
Provide an appropriate response.
- If f(x) (^) = 2x^3 - 5x (^) + 5, show that there is at least one value of c for which f(x) equals π.
Calculate the derivative of the function. Then find the value of the derivative as specified.
g(x) (^) = 3x^2 - 4x; g ′(3)
f(x) = x^2 + 7x - 2; f ′(0)
Given the graph of f, find any values of x at which f ′ is not defined.
Provide an appropriate response.
- If g(x) = 2f(x) + 3, find g ′(4) given that f ′ (4) =
- If g(x) (^) = - f(x) (^) - 3, find g ′(4) given that f ′(4) (^) =
Find the second derivative.
- y = 5x^2 + 8x + 4x-^3
Find the derivative of the function.
- y = 2 -^ 2x
(^2) + x 5 x
Suppose u and v are differentiable functions of x. Use the given values of the functions and their derivatives to find the value of the indicated derivative.
- u(2) = 9 , u ′(2) = 4 , v(2) = - 3 , v ′(2) = - 5. d dx
(3v - u) at x = 2
Solve the problem.
- Under standard conditions, molecules of a gas collide billions of times per second. If each molecule has diameter t, the average distance between collisions is given by L = 1 2 πt2n
where n, the volume density of the gas, is a constant. Find dL dt
Provide an appropriate response.
- Find d dx
x^3 - 2 x
by using the Quotient Rule
and by using the Product Rule. Show that your answers are equivalent.
Solve the problem.
- At time t ≥ 0, the velocity of a body moving along the s-axis is v = t2^ - 9t + 8. When is the body's velocity increasing?
The figure shows the velocity v or position s of a body moving along a coordinate line as a function of time t. Use the figure to answer the question.
- v (ft/sec)
1 2 3 4 5 6 7 8 9 10
5 4 3 2 1
1 2 3 4 5 6 7 8 9 10
5 4 3 2 1
t (sec)
When is the body moving backward?
Solve the problem.
- The size of a population of lions after t months is P = 100 (1 + 0.2t + 0.02t2). Find the growth rate when P (^) = 2500.
Find the derivative.
- s = t^4 cos t - 6t sin t - 6 cos t
Solve the problem.
Find the tangent to y = 2 - sin x at x = π.
Find an equation for the tangent to the curve at P.
Find the limit.
- lim x→π/
32 + sin(π sec x)
Provide an appropriate response.
- Find d^998 /dx^998 (sin x).
Given y = f(u) and u = g(x), find dy/dx = f′(g(x))g′(x).
- y (^) = 6 u^
, u (^) = 3x (^) - 7
- y = sin u, u = 9 x + 11
Write the function in the form y = f(u) and u = g(x). Then find dy/dx as a function of x.
y (^) = cos^4 x
y = cot( 8 x - 3 )
Find the derivative of the function.
- y = cos^5 (πt - 13)
Answer Key
Testname: STUDY GUIDE E
y = e-x; y = 2x
y = - ex; y = x
- No
- (-∞,∞)
- Notice that f(0) = 5 and f(1) = 2. As f is continuous on [0,1], the Intermediate Value Theorem implies that there is a number c such that f(c) = π.
- g ′(x) = 6x - 4; g ′(3) = 14
- f ′(x) = 2x + 7; f ′(0) = 7
- x (^) = - 2, 0, 2
- g ′(4) = 10
- g ′(4) = - 5
- 10 + 48x-^5
- dy dx = -^18 x^
x^
x
- (^) - 19
- dL dt
2 πt3n
- By the Quotient Rule, d dx
x^3 - 2 x
= x^ ·^ (3x
(^2) ) - (x (^3) - 2) · 1 x^
= 2x
x^
By the Product Rule, d dx
x3^ - 2 x
= d dx
((x3^ - 2) x-1) = (x3^ - 2)(-x-2) + x-1(3x2) = 2x + 2x-^2 = 2x
x^
2x3^ + 2 x^
= 2x
x
- t > 4.
- 0 < t < 4
- 140 lions/month
- ds dt
= - t4 sin t + 4t3 cos t - 6t cos t
y = x - π + 2
y = 2 x - π^2 4
- 9 cos ( 9 x + 11 )
- y (^) = u4; u (^) = cos x; dy dx =^ -^
4 cos3 x sin x
- y = cot u; u = 8x - 3; dy dx
= - 8 csc2(8x - 3)
Answer Key
Testname: STUDY GUIDE E
- dy dt
= - 5π cos4(πt - 13) sin(πt - 13)
4x
( x (^) - 8)-^5 - 8 x
- x (^) = y^2
-5 -4 -3 -2 -1 1 2 3 4 5 x
5 y 4 3 2 1
-5 -4 -3 -2 -1 1 2 3 4 5 x
5 y 4 3 2 1
Entire parabola, bottom to top (from fourth quadrant to origin to first quadrant)
- y = 1 8
x + 4
1 48
m/sec
