Let fx=-2x+4 and gx=-6x-7

Let f(x)=-2x+4 and g(x)=-6x-7. Find f(x) - g(x). Get the answers you need, now! Just put what they equal in parentheses, with a minus sign between them and simplify. f(x)-g(x) (-2x+4)-(-6x-7) -2x+4+6x+7 4x+11 Doesn't take a rocket scientist !

f (x), of the following functions. Show all your beautiful algebra. (a) f(x)=2x lim x2(x + 0)2. = −6x x4. = −6 x3. 3. For the following questions consider f(x)=4/x. 2 4. Let f(x) = √ x − 5. (a) Find the equation of the secant line that goes through the   4 − x2. = 1. 4 . 3. For which value of the constant c is the function f(x) Solution. f (x) = (2x + 3)(6x5 − 2x8)+(x2 + 3x)(30x4 − 16x7). f (1) = 5 · 4+4 · 14 = 76. 7. For f(x ) = 3. √ x5 + 6. 5√ x3 , find f (x). Solution. Rewriting f(x) = x. 5. 3 + 6x−3 4. REPORT. Solution. Let f(x) = 1 x2 + 1 . Then y = f (x) = lim h→0 f(x + h) − f(x) h. = lim. 29 Jan 2012 5(1)4 + 3(1)2 + 1. = 1. 9. 3. Find (f-1)/(x) using formula 2 f(x) = 2 x + 3 f/(x) = -. 2 4. 2x2. = -. 2 x2. 7. Find the derivative of f-1 by using formula 3. y = 5x3 + x - 7 ( x3 - 2x)ln(x)(ln(x3 - 2x) x. +. (3x2 - 2) ln(x) x3 - 2x. ) 43. Let f(x) = x4 + x3 + 1, 0 ≤ x ≤ 2. (b) Let g(x) = f-1(x) and define F(x) = f(2g(x)). Find an  g(x). (d) h(x) = f(x). 1 + f(x). Solution: (a) h (2) = 5f (2) − 4g (2) = 5(−2) − 4(7) = −38 Let f(x) = { x2 if x ≤ 2 mx + b if 2 < x. . Find the values of m and b that make f(x) differentiable everywhere. Solution: (a) Clearly, f (x)=2x for x < 2, f (x) = m for x > 2 . 11π. 6. + 2πn, where n is any integer. 5. Find lim x→0 sin 4x sin 6x . Solution:. That would mean that f(2) and f(4) both equal 3, and one-to-one functions can't assign two Below is the graph of g : R ! R where g(x) = x3. Let f : R ! R be the function defined by f(x)=2x + 2, and let If f(x + 7) = 1, then x +7= f 1(1). • If f 1(0)  $f=2x+3,\:g=-x^2+5,\:f\left(x\right)\:\circ\:g\left(x\right):\quad-2x^2+13$ f =2 x +3, $\mathrm{For}\:f=2x+3\:\mathrm{substitute}\:x\:\mathrm{with}\:g\left(x\right)=-x^2  Complete the table below and plot the graphs of \(f(x)\) and \(g(x)\) on the same \[f(x) = 2x^{3} - 5x^{2} - 14x + 8 \qquad \qquad g(x) = -2x^{3} + 5x^{2} + 14x - 8\] \begin{align*} \text{Let } f(x) &= x^{3} - \frac{5}{4}x^{2} - \frac{7}{4}x + \frac{1}{2} \\ f (-1) x(x - 4)(x+1)\\ \therefore x=-1, \enspace & x = 0 \text{ or } x = 4 \end{align*}.

11x-37=x+23 10p-26=p-44 -13x+5=4x-6 -7x+13=4x-7 11x+5=17 -11x+13=-42 - 15y+26=-19 5x-13=9x+15 6q-14=11q-19 4(-2x-5)=29 12(2-5x)=(3x-1) 

the third-degree Taylor polynomial for g(x) = e* f(x) about x = 0. -4 + 2x -. 2. 3*. T 10%. (c) Let Q(x) denote the Taylor polynomial of degree n for h about 35+7- 9*. -. This series converges by the Alternating Series Test. Therefore, the interval   30 Dec 2013 Given the functions f : x 3x 7 and g : x 4 x 1 , find (a) f g(1) 1. f(x) = 2x + 1 f –1(7) = k Let when y = –1 f (x) = f –1(7) = = f(k) = 7 x = 2y + 1 x – 1 Given the functions g : x 2x - 1 and gf : x 6x + 7, find the function g,. fg  For example, a2a3 = a5. Multiply 3x2· 4x5· 2x a), 5x2· 6x4 = 30x6, b), 7x3· 8x 6 = 56x9 2· 2 = 4 is not correct here. Within the parentheses there are three factors: 2, x3, and y4. Problem 7. f) x2· x2 = x4. Rule 1. g) x2· y3 Not possible. Let f(x) = -2x + 4 and g(x)= -6x - 7. Find f(x) - g(x ... 9 minutes ago If the mean weight of 4 backfield members on the football team is 212 lb and the mean weight of the 7 other players is 203 lb, what is the mean weigh

29 Jan 2012 5(1)4 + 3(1)2 + 1. = 1. 9. 3. Find (f-1)/(x) using formula 2 f(x) = 2 x + 3 f/(x) = -. 2 4. 2x2. = -. 2 x2. 7. Find the derivative of f-1 by using formula 3. y = 5x3 + x - 7 ( x3 - 2x)ln(x)(ln(x3 - 2x) x. +. (3x2 - 2) ln(x) x3 - 2x. ) 43. Let f(x) = x4 + x3 + 1, 0 ≤ x ≤ 2. (b) Let g(x) = f-1(x) and define F(x) = f(2g(x)). Find an 

f (x), of the following functions. Show all your beautiful algebra. (a) f(x)=2x lim x2(x + 0)2. = −6x x4. = −6 x3. 3. For the following questions consider f(x)=4/x. 2 4. Let f(x) = √ x − 5. (a) Find the equation of the secant line that goes through the   4 − x2. = 1. 4 . 3. For which value of the constant c is the function f(x) Solution. f (x) = (2x + 3)(6x5 − 2x8)+(x2 + 3x)(30x4 − 16x7). f (1) = 5 · 4+4 · 14 = 76. 7. For f(x ) = 3. √ x5 + 6. 5√ x3 , find f (x). Solution. Rewriting f(x) = x. 5. 3 + 6x−3 4. REPORT. Solution. Let f(x) = 1 x2 + 1 . Then y = f (x) = lim h→0 f(x + h) − f(x) h. = lim. 29 Jan 2012 5(1)4 + 3(1)2 + 1. = 1. 9. 3. Find (f-1)/(x) using formula 2 f(x) = 2 x + 3 f/(x) = -. 2 4. 2x2. = -. 2 x2. 7. Find the derivative of f-1 by using formula 3. y = 5x3 + x - 7 ( x3 - 2x)ln(x)(ln(x3 - 2x) x. +. (3x2 - 2) ln(x) x3 - 2x. ) 43. Let f(x) = x4 + x3 + 1, 0 ≤ x ≤ 2. (b) Let g(x) = f-1(x) and define F(x) = f(2g(x)). Find an 

g(x). (d) h(x) = f(x). 1 + f(x). Solution: (a) h (2) = 5f (2) − 4g (2) = 5(−2) − 4(7) = −38 Let f(x) = { x2 if x ≤ 2 mx + b if 2 < x. . Find the values of m and b that make f(x) differentiable everywhere. Solution: (a) Clearly, f (x)=2x for x < 2, f (x) = m for x > 2 . 11π. 6. + 2πn, where n is any integer. 5. Find lim x→0 sin 4x sin 6x . Solution:.

So the tangent line is y − 1/2 = −(1/4)(x − 2). (2) Let f(x) = { x2 if x ≤ 2 mx + b if x > 2 1. 7 . (c) lim x→0 sin(3x) tan(2x) lim x→0 sin(3x) tan(2x). = lim x→0. (sin(3x) sin(2x). ) cos(2x) By Chain Rule, F (x)=3f2(x)f (x) and G (x) = f (x3)(3x2). So F (1 )  I. The ability to set up and simplify difference quotients is essential for calculus 6x 3h 4. +. −. B. = [steps are identical to A, but using “a” for”x”] f(a h) f(a) h. +. − + . VII. Examples using radical functions. A. Given f(x) x. = 1. = . f(x h) f(x) h. + −. x h +. VIII. Examples using rational functions. A. Given. 1 g(x) x. = 1. = g(x h) g(x) h. 9 Oct 2014 Understanding function notation. Need more help? Mr. Dwyer is available for 1- on-1 tutoring online. Details at 

15 Mar 2011 Help your child succeed in math at https://www.patreon.com/tucsonmathdoc Given f(x)=3x+2 and g(x)=2x^2-1 find the following and state if te 

As an example, let f(x) = x2 + 1 and g(x)=2x − 4. Then (b) (f ◦ g)(x) = f(g(x)) = f(x + 4) = 2(x +4)+3=2x + 11. (c) (g ◦ f)(x) = g(f(x)) = g(2x + 3) = (2x +3)+4=2x + 7. 2x + 1. 3x − 4. = 2. ⇔ 2x + 1 = 2(3x − 4). ⇔ 2x + 1 = 6x − 8. ⇔ 4x = 9. ⇔ x = 9. 4.

the third-degree Taylor polynomial for g(x) = e* f(x) about x = 0. -4 + 2x -. 2. 3*. T 10%. (c) Let Q(x) denote the Taylor polynomial of degree n for h about 35+7- 9*. -. This series converges by the Alternating Series Test. Therefore, the interval   30 Dec 2013 Given the functions f : x 3x 7 and g : x 4 x 1 , find (a) f g(1) 1. f(x) = 2x + 1 f –1(7) = k Let when y = –1 f (x) = f –1(7) = = f(k) = 7 x = 2y + 1 x – 1 Given the functions g : x 2x - 1 and gf : x 6x + 7, find the function g,. fg  For example, a2a3 = a5. Multiply 3x2· 4x5· 2x a), 5x2· 6x4 = 30x6, b), 7x3· 8x 6 = 56x9 2· 2 = 4 is not correct here. Within the parentheses there are three factors: 2, x3, and y4. Problem 7. f) x2· x2 = x4. Rule 1. g) x2· y3 Not possible. Let f(x) = -2x + 4 and g(x)= -6x - 7. Find f(x) - g(x ... 9 minutes ago If the mean weight of 4 backfield members on the football team is 212 lb and the mean weight of the 7 other players is 203 lb, what is the mean weigh please help! I dont no how to do this - Algebra