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DochEvi [55]
3 years ago
15

Erify that f has an inverse. then use the function f and the given real number a to find (f −1)'(a). (hint: see example 1. if an

answer does not exist, enter dne.) f(x) = x + 2 x − 1 , x > 1 a = 2

Mathematics
1 answer:
zhannawk [14.2K]3 years ago
8 0

The function

... f(x) = (x+2)/(x-1) = 1 + 3/(x-1)

is symmetrical about the line y=x, hence is its own inverse.


We can evaluate the desired derivative directly.

... f'(x) -3/(x-1)²

so f'(2) is

... f'(2) = -3/(2-1)²


(f^{-1})(2)=-3

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How do I do this activity?
d1i1m1o1n [39]

In the X column put 1, 2, and 3

for the y column, plug in the 1, 2, or 3 into the X

for example, a) y=5x+3  

x= 1, 2, 3                 y=5(1)+3 = 8

y= 8, 13, 18

7 0
3 years ago
1 + tanx / 1 + cotx =2
Lera25 [3.4K]

Answer:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

Step-by-step explanation:

Solve for x:

1 + cot(x) + tan(x) = 2

Multiply both sides of 1 + cot(x) + tan(x) = 2 by tan(x):

1 + tan(x) + tan^2(x) = 2 tan(x)

Subtract 2 tan(x) from both sides:

1 - tan(x) + tan^2(x) = 0

Subtract 1 from both sides:

tan^2(x) - tan(x) = -1

Add 1/4 to both sides:

1/4 - tan(x) + tan^2(x) = -3/4

Write the left hand side as a square:

(tan(x) - 1/2)^2 = -3/4

Take the square root of both sides:

tan(x) - 1/2 = (i sqrt(3))/2 or tan(x) - 1/2 = -(i sqrt(3))/2

Add 1/2 to both sides:

tan(x) = 1/2 + (i sqrt(3))/2 or tan(x) - 1/2 = -(i sqrt(3))/2

Take the inverse tangent of both sides:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or tan(x) - 1/2 = -(i sqrt(3))/2

Add 1/2 to both sides:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or tan(x) = 1/2 - (i sqrt(3))/2

Take the inverse tangent of both sides:

Answer:  x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

4 0
3 years ago
HELP!!!!!!!!!! WILL GIVE BRAINLY!!!!!!!
atroni [7]

Answer:

141in^{2}

Step-by-step explanation:

Area of a trapezium = \frac{h}{2}(a + b)

h = height

a and b = upper and and lower lengths

Area of a trapezium = \frac{9}{2}(10 + 16)

 =  \frac{9}{2} × 26

 = 117in^{2}

Area of a triangle = \frac{1}{2} × base × height

 = \frac{1}{2} × 6 × 8

 =  \frac{1}{2} × 48

 = 24in^{2}

Total area = Sum of the areas of both the shapes

 = 117 + 24

 = 141in^{2}

5 0
2 years ago
ANYBODY?!?!! HELPPP PLEASEEE!
Zielflug [23.3K]

egyuyeyuehebdfhvgebfjhaifhldan dhsgojkna jbhope

6 0
3 years ago
Mei has 8 jars of soup. Each jar contains 300 milliliters of soup. What is the smallest pot mei can use to heat all the soup? a.
NeX [460]

Answer:

Option B. 3 liters

Step-by-step explanation:

First, we shall determine the total volume of soup that Mei have. This is illustrated below:

Mei have 8 jars of soup containing 300mL each. Therefore, she has a total of = 8 x 300mL = 2400mL.

Next, we shall convert 2400mL to litres (L). This can be obtain as follow:

1000mL = 1L

Therefore, 2400mL = 2400/1000 = 2.4L

Mei have 2.4L of soup.

Since the total volume of the soup that Mei have is 2.4L, she will be needing a minimum of 3L pot to heat up all the soup.

3 0
3 years ago
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