gevaughn600

18.11.2020 •

Mathematics

Consider these functions: f(x)=3x-7
g(x)=x+1/x-1
what is the value of f(g(3))
a. -1
b. 0
c. 3
d. 5

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Ответ:

perrintaylor01

15.04.2023

Mathematics

see both proofs below

Step-by-step explanation:

Use Difference Identity: tan (A - B) = (tan A - tan B)/(1 + tanA · tanB)

Use Unit Circle to evaluate: tan (π/4) = 1

Use Tangent Identity: tanA = (sinA)/(cosA)

Use Half-Angle Identities:

$\cos \dfrac{A}{2}=\sqrt{\dfrac{1+\cos A}{2}}\\\\\\\sin \dfrac{A}{2}=\sqrt{\dfrac{1-\cos A}{2}}$

Part 1 Proof LHS → Middle

LHS: $\tan\bigg(\dfrac{\pi}{4}-\dfrac{A}{2}\bigg)$

Difference Identity: $\dfrac{\tan (\frac{\pi}{4})-\tan(\frac{A}{2})}{1+\tan(\frac{\pi}{4})\cdot \tan(\frac{A}{2})}$

Unit Circle: $\dfrac{1-\tan(\frac{A}{2})}{1+ \tan(\frac{A}{2})}$

$\text{Tangent Identity:}\qquad \qquad \dfrac{\frac{\cos\frac{A}{2}-\sin\frac{A}{2}}{\cos\frac{A}{2}}}{\frac{\cos\frac{A}{2}+\sin\frac{A}{2}}{\cos\frac{A}{2}}}$

Simplify: $\dfrac{\cos\frac{A}{2}-\sin\frac{A}{2}}{\cos\frac{A}{2}+\sin\frac{A}{2}}$

LHS = Middle: $\dfrac{\cos\frac{A}{2}-\sin\frac{A}{2}}{\cos\frac{A}{2}+\sin\frac{A}{2}}=\dfrac{\cos\frac{A}{2}-\sin\frac{A}{2}}{\cos\frac{A}{2}+\sin\frac{A}{2}}\qquad \checkmark$

Part 2 Proof Middle → RHS

Middle: $\dfrac{\cos\frac{A}{2}-\sin\frac{A}{2}}{\cos\frac{A}{2}+\sin\frac{A}{2}}$

$\text{Half-Angle Identity:}\qquad \qquad \dfrac{\sqrt{\frac{1+\cos A}{2}}-\sqrt{\frac{1-\cos A}{2}}}{\sqrt{\frac{1+\cos A}{2}}+\sqrt{\frac{1-\cos A}{2}}}$

Simplify: $\dfrac{\sqrt{1+\cos A}-\sqrt{1-\cos A}}{\sqrt{1+\cos A}+\sqrt{1-\cos A}}$

Rationalize Denominator: $\dfrac{\sqrt{1+\cos A}-\sqrt{1-\cos A}}{\sqrt{1+\cos A}+\sqrt{1-\cos A}}\bigg(\dfrac{\sqrt{1+\cos A}-\sqrt{1-\cos A}}{\sqrt{1+\cos A}-\sqrt{1-\cos A}}\bigg)$

$=\dfrac{1+\cos A-2\sqrt{1-\cos^2 A}+1-\cos A}{1+\cos A-(1-\cos A)}$

Simplify: $\dfrac{2-2\sqrt{1-\cos^2 A}}{2\cos A}$

$=\dfrac{2-2\sqrt{sin^2 A}}{2\cos A}$

$= \dfrac{2-2\sin A}{2\cos A}$

Factor: $\dfrac{2(1-\sin A)}{2(\cos A)}$

Simplify: $\dfrac{1-\sin A}{\cos A}$

Expand: $\dfrac{1-\sin A}{\cos A}\bigg(\dfrac{1+\sin A}{1+\sin A}\bigg)$

$=\dfrac{1-\sin^2 A}{\cos A(1+\sin A)}$

Simplify: $\dfrac{\cos^2 A}{\cos A(1+\sin A)}$

$=\dfrac{\cos A}{1+\sin A}$

Middle = RHS: $\dfrac{\cos A}{1+\sin A}=\dfrac{\cos A}{1+\sin A}\qquad \checkmark$

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Consider these functions: f(x)=3x-7 g(x)=x+1/x-1 what is the value of f(g(3))a. -1b. 0c. 3d. 5