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3 years ago

Determine whether each pair of functions are inverse functions. Write yes or no.

Mathematics

1 answer:

Tatiana [17]3 years ago

6 0

Answer: No

Step-by-step explanation:

There are two ways to solve this. The first way is to calculate g(f(x)) (or f(g(x))). If f and g are inverse functions f(g(x)) = g(f(x)) = x.

g(f(x)) = (1/4)(2sqrt(x - 5))^2 - 5

= (1/4)*2^2*sqrt(x - 5)^2 - 5

= (1/4)*4*(x - 5) - 5

= x - 5 - 5

= x - 10

g(f(x)) = x - 10 =/= x, so the two functions are not inverse functions.

You can also solve this by using a test value a and calculating g(f(a)). If f and g are inverse functions, the result will be a.

For example, let a = 6.

f(6) = 2*sqrt(6 - 5)

= 2*sqrt(1)

= 2*1

= 1

g(f(6)) = g(1)

= (1/4)*1^2 - 5

= (1/4)*1 - 5

= 1/4 - 5

= -4 3/4

=/= 6

Therefore f and g are not inverse functions

Please note that this technique does not work in the other direction. If g(f(a)) = a, this does NOT prove that f and g are inverse functions.

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Hello from MrBillDoesMath!

Answer:

The first marker is labeled 208708; the second (consecutive) one 208709

Discussion:

Call the first mile marker "m". Then

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2 years ago

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3 years ago

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Which equation has the solutions x = -3 ± √3i/2 ?

Maurinko [17]

Answer:Answer is option C : [ $x^{2}$ + 3x + 3 ] =0

Note: None of options matches with given question.

instead of "-3" , there should be "- $\frac{3}{2}$ ".

Step-by-step explanation:

Note: None of options matches with given question.

instead of "-3" , there should be " $\frac{3}{2}$ ".

Here, First thing you have to observe the nature of roots.

∴ x = - $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i and x = - $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$

∴ [ x+( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i) ][ x+( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i) ]=0

∴ [ $x^{2}$ + x( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i)+ x( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i) + ( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i)( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i) ]=0

∴ [ $x^{2}$ + $\frac{3}{2}$ x + $\frac{\sqrt{3}}{2}$ ix + $\frac{3}{2}$ x - $\frac{\sqrt{3}}{2}$ ix + (3- $\frac{\sqrt{3}}{2}$ i)(3+ $\frac{\sqrt{3}}{2}$ i) ] =0

∴ [ $x^{2}$ + 3x + ( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i)( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i) ] =0

∴ [ $x^{2}$ + 3x + $\frac{9}{4}$ - ( $\frac{\sqrt{3}}{2}$ i)( $\frac{\sqrt{3}}{2}$ i) ] =0

∴ [ $x^{2}$ + 3x + $\frac{9}{4}$ - ( $\frac{3}{4}$ ) $i^{2}$ ] =0

∴ [ $x^{2}$ + 3x + $\frac{9}{4}$ + ( $\frac{3}{4}$ ) ] =0

∴ [ $x^{2}$ + 3x + $\frac{12}{4}$ ] =0

∴ [ $x^{2}$ + 3x + 3 ] =0

Thus, Answer is option C : [ $x^{2}$ + 3x + 3 ] =0

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