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AVprozaik [17]
3 years ago
13

If ​f(x) =

class="latex-formula"> and ​g(x) = x cubed plus 8​, simplify the expressions (f circle g )(2 )​, (f circle f )(25 )​, (g circle f )(x )​, and (f circle g )(x ).
Mathematics
1 answer:
kogti [31]3 years ago
8 0

Answer:

Step-by-step explanation:

Given are two functions f and g defined as

f(x) =\sqrt{x} \\g(x) = x^3+8

we have to find the composition of functions in all orders

1) a)(fog)(x)=f{g(x)}=f(x^3+8) =\sqrt{x^3+8} \\\\(fog)(2) = =4,-4

b) (fof)(x) = f{f(x)}=f(\sqrt{x} {=\sqrt[4]{x} \\\\When x=25, ans= \sqrt{5}

c) (gof)(x) = g(\sqrt{x}) = (\sqrt{x})^3+3 = x^{\frac{3}{2} } +8\\\\d)(fog)(x)=f{g(x)}=f(x^3+8)

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3 years ago
Solve the equation cos (x/2) = cos x + 1. what are the solutions on the interval 0° ≤ x < 360°?
Sedbober [7]

Answer:

Step-by-step explanation:

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cos (x/2)=2cos ²(x/2)

2 cos²(x/2)-cos (x/2)=0

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x/2=2nπ+π/2,2nπ+3π/2

x=4nπ+π,4nπ+3π

n=0,1,2,...

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so x=180°

or

2cos(x/2)-1=0

cos (x/2)=1/2=cos60,cos (360-60)=cos 60,cos 300=cos (360n+60),cos (360n+300)

x/2=360n+60,360n+300

x=720n+120,720n+300

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x=120,300,840,1020,...

only 120° and 300° ∈[0,360°]

Hence x=120°,180°,300°

7 0
2 years ago
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2a+8a
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Answer:

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3 years ago
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