Transformations of Functions
points from the graph of ( ) 3. f x = x and multiplying each of the x-values by -1. Page 11. Example 3 (Continued):. Step 3: Thus we have obtained the graph of
Limits Quiz Review
3. 4. 3. F. 0. -10. 2. 0. KK. 1. #. ?. 4 -2. 10. 8. 6. +6. -8. 12= f(x)=. 6. 3. (x-3)(z - 1). 1. 8 10. 4. Domain: [-200) lim f(x) x?1- lim f(x) x?3.
DÉRIVATION
y = 7 x ? 2. ( )+ 9 y = 7x ? 5. Une équation de tangente à la courbe représentative de f au point A de la courbe d'abscisse 2 est y = 7x ? 5. 3) Formules
Week 3 Quiz: Differential Calculus: The Derivative and Rules of
As x approaches 3 from below and from above the value of the function f(x) approaches f(3) = 6. Thus the limit limx?3f(x) = 6. Question 2: Find limx?2f(x): f
Cambridge IGCSE
(f) (i) The equation f(x) = x2 can be written as x3 + px2 + q = 0. Show that p = ?1 and q = ?20. [2]. (ii) On the grid opposite draw the graph of y = x2
Numerical Mathematical Analysis
3. Rootfinding. Calculating the roots of an equation f(x)=0. (7.1) is a common problem in applied mathematics. We will explore some simple numerical methods
SOLUTIONS Problem 1. Find the critical points of the function f(x y
The three critical points are. (00)
College Algebra Section 4.3
https://www.sccollege.edu/Departments/MATH/Documents/Math%20140/04-03-044.pdf
FONCTION DERIVÉE
Soit la fonction f définie sur R par f (x) = x3 +. 9. 2 x2 ?12x +5. 1) Etudier les variations de f et dresser le tableau de variation. 2) Dans repère
1.6 Graphs of Functions
1 plot the points and connect the dots in a somewhat pleasing fashion to get the graph below on the right. x f(x) (x
Transformation of Functions Worksheet - Loyola University Chicago
Unit 1: Lesson 3 Transformations of Graphs Hour_____ Graph the following functions without using technology Feel free to use a graphing calculator to check your answer but you should be able to look at the function and apply what you learned in the lesson to move its parent function
Solutions to HW5 Problem 31 - IUPUI
(a) To ?nd E[X] we ?rst ?nd the PDF by di?erentiating the above CDF fX (x) = ˆ 1/2 0 ? x ? 2 0 otherwise (2) The expected value is then E[X] = Z 2 0 x 2 dx = 1 (3) (b) E X2 = Z 2 0 x2 2 dx = 8/6 = 4/3 (4) Var[X] = E X2 ?E [X]2 = 4/3 ?1 = 1/3 (5) Problem 3 3 4 • The probability density function of random variable Y is fY (y
Derivatives - limit definition - Santa Barbara City College
(b) fx x x( ) 2 7= +2 (Use your result from the second example on page 2 to help ) (c) fx x x( ) 4 6= ? 3 (Use the second example on page 3 as a guide ) Check your answers at the end of this document
Searches related to fx = x^3 PDF
Example 1 Consider random variables XY with pdf f(xy) such that f(x;y) = 8
Is f(x) = |x - 3| negative?
The value of f (x) = |x - 3| is never negative,. For value of x < 3, the function f (x) = 3 - x which is positive. The range of the function is therefore the set [0, oo) .
What does f(x)=3 mean?
What does f (x)==3 mean? It means that you have a function that every time you input a value of x gives you 3. If x = 45 then y = 3 ...if x = ? 1.234 then, again, y = 3 and so on!
What is the domain of the function f(x) = 3x?
[Solved] What is the domain of the function f (x) = 3x? What is the domain of the function f (x) = 3 x? The domain is the set of all possible value of x which have a finite value of f (x). Mistake Points The range of the given function will be from (0,?). 0, when x = -?, and ? when x = ?.
Is f(x) = 3x2 an exponential function?
Is f (x) = 3x2 an exponential function? No, it is a quadratic function. An exponential function has the variable as the exponent. 3x would have been an example of an exponential function. In this function, the exponent is not variable, it is 2.
SGPE Summer School 2016
Limits
Question 1:Find limx!3f(x):
f(x) =x29x3 (A) +1 (B) -6 (C) 6 (D) Does not exist! (E) None of the above Answer:(C) Note the the functionf(x) =x29x3=(x3)(x+3)x3=x+ 3 is actually a line. However it is important to note the this function isundenedatx= 3. Why?x= 3 requires dividing by zero (which is inadmissible). Asxapproaches 3 from below and from above, the value of the functionf(x) approachesf(3) = 6. Thus the limit limx!3f(x) = 6.Question 2:Find limx!2f(x):
f(x) = 1776 (A) +1 (B) 1770 (C)1 (D) Does not exist! (E) None of the above Answer:(E) The limit of any constant function at any point, sayf(x) =C, whereCis an arbitrary constant, is simplyC. Thus the correct answer is limx!2f(x) = 1776.Question 3:Find limx!4f(x):
f(x) =ax2+bx+c (A) +1 (B) 16a + 4b + c (C)1 (D) Does not exist! (E) None of the above 1Answer:(B) Applying the rules of limits:
lim x!4ax2+bx+c= limx!4ax2+ limx!4bx+ limx!4c =a[limx!4x]2+blimx!4x+c = 16a+ 4b+cQuestion 4:Find the limits in each case:
(i) lim x!0x 2jxj (ii) lim x!32x+34x9 (iii) lim x!6x23xx+3
Answer:(i) limx!0x
2jxj= limx!0(jxj)2jxj= limx!0jxj= 0
(ii) limx!32x+34x9=23+3439= 3 (iii) limx!6x23xx+3=62366+3
= 2 Question 5:Show that limx!0sinx= 0 (Hint:xsinxxfor allx0.) Answer:Given hint and squeeze theorem we have limx!0x= 0limx!0sinx0 = limx!0xhence, lim x to0sinx= 0Question 6:Show that limx!0xsin(1x
) = 0 Answer:Note rst that for any real numbertwe have1sint1 so1sin(1x )1. Therefore, xxsin(1x )xand by squeeze theorem limx!0xsin1x = 0.Continuity and Dierentiability
Question 7:Which of the following functions areNOTeverywhere continuous: (A)f(x) =x24x+2 (B)f(x) = (x+ 3)4 (C)f(x) = 1066 (D)f(x) =mx+b (E) None of the above Answer:(A) Remember that, informally at least, acontinuousfunction is one in which there are nobreaks its curve. A continuous function can be drawn without lifting your pencil from the paper. More
formally, a functionf(x) iscontinuousat the pointx=aif and only if:1.f(x) is dened at the pointx=a,
2. the limit lim
x!af(x) exists,3. lim
x!af(x) =f(a) The functionf(x) =x24x+2is not everywhere continuous because the function is not dened at the point x=2. It is worth noting that limx!2f(x) does in fact exist!The existence of a limit at a point does not guarantee that the function is continuous at that point! 2 Question 8:Which of the following functions are continuous: (A)f(x) =jxj (B)f(x) =3x <4 12 x+ 3x4 (C)f(x) =1x (D)f(x) =lnx x <0 0x= 0 (E) None of the above Answer:(A) The absolute value functionf(x) =jxjis dened as: f(x) =x x0 x x <0 Does this function satisfy the requirements for continuity? Yes! The critical point to check isx= 0. Note that the function is dened atx= 0; the limx!0f(x) exists; and that limx!0f(x) = 0 =f(0). Question 9:Which of the following functions areNOTdierentiable: (A)f(x) =jxj (B)f(x) = (x+ 3)4 (C)f(x) = 1066 (D)f(x) =mx+b (E) None of the above Answer:(A) Remember that continuity is anecessarycondition for dierentiability (i.e., every dier- entiable function is continuous), but continuity is not asucientcondition to ensure dierentiability(i.e., not every continuous function is dierentiable). Case in point isf(x) =jxj. This function is in
fact continuous (see previous question). It is not however dierentiable at the pointx= 0. Why? The pointx= 0 is a cusp (or kink). There are an innite number of lines that could be tangent to the functionf(x) =jxjat the pointx= 0, and thus the derivative off(x) would have an innite number of possible values.Question 10:Is function
f(x) =0 :x= 0 xsin(1=x) :x6= 0 continuous at point 0?Answer:Note thatfis continuous at a pointaif
lim x!af(x) =f(limx!ax):In this case, we takea= 0 and
lim x!0f(x) = limx!0xsin(1=x) = 0 by question 6. Moreover, f(limx!0x) =f(0) = 0 thus,fis continuous at 0. 3Derivatives
Question 11:Find the derivative of the following function: f(x) = 1963 (A) +1 (B) 1963 (C)1 (D) 0 (E) None of the above Answer:(D) The derivative of a constant function is always zero. Question 12:Find the derivative of the following function: f(x) =x2+ 6x+ 9 (A)f0(x) = 2x+ 6 + 9 (B)f0(x) =x2+ 6 (C)f0(x) = 2x+ 6 (D)f0(x) = 2x (E) None of the aboveAnswer:(C) Remember that 1) the derivative of a sum of functions is simply the sum of the derivatives
of each of the functions, and 2) the power rule for derivatives says that iff(x) =kxn, thenf0(x) = nkx n1. Thusf0(x) = 2x21+ 6x11+ 0 = 2x+ 6. Question 13:Find the derivative of the following function: f(x) =x12 (A)f0(x) =12 px (B)f0(x) =1px (C)f0(x) =12 px (D)f0(x) =px (E) None of the above Answer:(C) Remember that the power rule for derivatives works with fractional exponents as well!Thusf0(x) =12
x12 1=12 x12 =12 px Question 14:Find the derivative of the following function: f(x) = 5x2(x+ 47) (A)f0(x) = 15x2+ 470x (B)f0(x) = 5x2+ 470x (C)f0(x) = 10x (D)f0(x) = 15x2470x 4 (E) None of the above Answer:(A) Ideally, you would solve this problem by applying the product rule. Setg(x) = 5x2and h(x) = (x+ 47), thenf(x) =g(x)h(x). Apply the product rule: f0(x) =g0(x)h(x) +g(x)h0(x)
= 10x(x+ 47) + 5x2(1) = 10x2+ 470x+ 5x2 = 15x2+ 470x Question 15:Find the derivative of the following function: f(x) =5x2x+ 47 (A)f0(x) =5x2470x(x+47)2 (B)f0(x) =10x2+470x(x+47) (C)f0(x) = 10x (D)f0(x) =5x2+470(x+47)2 (E) None of the above Answer:(A) Ideally, you would solve this problem by applying the quotient rule. Setg(x) = 5x2and h(x) = (x+ 47), thenf(x) =g(x)h(x). Apply the quotient rule: f0(x) =g0(x)h(x)g(x)h0(x)h(x)2
10x(x+ 47)5x2(1)(x+ 47)2
10x2+ 470x5x2(x+ 47)2
5x2+ 470x(x+ 47)2
Question 16:Find the derivative of the following function: f(x) = 5(x+ 47)2 (A)f0(x) = 15x2+ 470x (B)f0(x) = 10x470 (C)f0(x) = 10x+ 470 (D)f0(x) = 15x2470x (E) None of the above Answer:(C) Ideally, you would solve this problem by applying the chain rule. Setg(h) = 5h2and h(x) = (x+ 47), thenf(x) =g(h(x)). Apply the chain rule: f0(x) =g0(h)h0(x)
= 10hquotesdbs_dbs26.pdfusesText_32[PDF] f(x)=2
[PDF] f(x)=x+1
[PDF] f'(x) dérivé
[PDF] f(x)=x^4
[PDF] f(x)=3
[PDF] livre mécanique appliquée pdf
[PDF] mécanique appliquée définition
[PDF] mécanique appliquée cours et exercices corrigés pdf
[PDF] mecanique appliquée bac pro
[PDF] pdf mecanique general
[PDF] mécanique appliquée et construction
[PDF] z+1/z-1 imaginaire pur
[PDF] z+1/z-1=2i
[PDF] questions ? poser lors dun audit interne
