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3241004551 [841]
3 years ago
14

If h(x)=(fog)(x) and h(x) = sqrt x+5, find g(x) if f(x) = sqrtx+2

Mathematics
1 answer:
Katen [24]3 years ago
3 0

Answer: The value of g(x) is x+3.

Step-by-step explanation:

Since we have given that

h(x)=f\circ g(x)

Here,

h(x)=\sqrt{x+5}\\\\and\\\\f(x)=\sqrt{x+2}

We need to find the value of g(x):

f(x)=\sqrt{x+2}\\\\f(g(x))=\sqrt{g(x)+2}\\\\h(x)=\sqrt{g(x)+2}\\\\\sqrt{x+5}=\sqrt{g(x)+2}\\\\x+5=g(x)+2\\\\g(x)=x+5-2=x+3

Hence, the value of g(x) is x+3.

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Answer:

If a+b+c=1,

a

2

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b

2

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2

,

a

3

+

b

3

+

c

3

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3

then find the value of

a

4

+

b

4

+

c

4

=

?

we know

2

(

a

b

+

b

c

+

c

a

)

=

(

a

+

b

+

c

)

2

−

(

a

2

+

b

2

+

c

2

)

⇒

2

(

a

b

+

b

c

+

c

a

)

=

1

2

−

2

=

−

1

⇒

a

b

+

b

c

+

c

a

=

−

1

2

given

a

3

+

b

3

+

c

3

=

3

⇒

a

3

+

b

3

+

c

3

−

3

a

b

c

+

3

a

b

c

=

3

⇒

(

a

+

b

+

c

)

(

a

2

+

b

2

+

c

2

−

a

b

−

b

c

−

c

a

)

+

3

a

b

c

=

3

⇒

(

a

+

b

+

c

)

(

a

2

+

b

2

+

c

2

−

(

a

b

+

b

c

+

c

a

)

+

3

a

b

c

=

3

⇒

(

1

×

(

2

−

(

−

1

2

)

+

3

a

b

c

)

)

=

3

⇒

(

2

+

1

2

)

+

3

a

b

c

=

3

⇒

3

a

b

c

=

3

−

5

2

=

1

2

⇒

a

b

c

=

1

6

Now

(

a

2

b

2

+

b

2

c

2

+

c

2

a

2

)

=

(

a

b

+

b

c

+

c

a

)

2

−

2

a

b

2

c

−

2

b

c

2

a

−

2

c

a

2

b

=

(

a

b

+

b

c

+

c

a

)

2

−

2

a

b

c

(

b

+

c

+

a

)

=

(

−

1

2

)

2

−

2

×

1

6

×

1

=

1

4

−

1

3

=

−

1

12

Now

a

4

+

b

4

+

c

4

=

(

a

2

+

b

2

+

c

2

)

2

−

2

(

a

2

b

2

+

b

2

c

2

+

c

2

a

2

)

=

2

2

−

2

×

(

−

1

12

)

=

4

+

1

6

=

4

1

6

Extension

a

5

+

b

5

+

c

5

=

(

a

3

+

b

3

+

c

3

)

(

a

2

+

b

2

+

c

2

)

−

[

a

3

(

b

2

+

c

2

)

+

b

3

(

c

2

+

a

2

)

+

c

3

(

a

2

+

c

2

)

]

=

3

⋅

2

−

[

a

3

(

b

2

+

c

2

)

+

b

3

(

c

2

+

a

2

)

+

c

3

(

a

2

+

b

2

)

]

Now

a

3

(

b

2

+

c

2

)

+

b

3

(

c

2

+

a

2

)

+

c

3

(

a

2

+

b

2

)

=

a

2

b

2

(

a

+

b

)

+

b

2

c

2

(

b

+

c

)

+

c

2

a

2

(

a

+

c

)

=

a

2

b

2

(

1

−

c

)

+

b

2

c

2

(

1

−

a

)

+

c

2

a

2

(

1

−

b

)

=

a

2

b

2

+

b

2

c

2

+

c

2

a

2

−

(

a

2

b

2

c

+

b

2

c

2

a

+

c

2

a

2

b

)

=

−

1

12

−

a

b

c

(

a

b

+

b

c

+

c

a

)

=

−

1

12

−

1

6

⋅

(

−

1

2

)

=

0

So

a

5

+

b

5

+

c

5

=

6

−

0

=

6

Step-by-step explanation:

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