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andre [41]
3 years ago
8

If f(x)=3^x+10 and g(x)=2x-4, find (f-g)(x)

Mathematics
1 answer:
nataly862011 [7]3 years ago
6 0
<h3>Answer:   3^x - 2x + 14</h3>

=====================================

Work Shown:

(f - g)(x) = f(x) - g(x)

(f - g)(x) = ( f(x) ) - ( g(x) )

(f - g)(x) = ( 3^x + 10 ) - ( 2x-4 )

(f - g)(x) = 3^x + 10 - 2x+4

(f - g)(x) = 3^x - 2x + (10+4)

(f - g)(x) = 3^x - 2x + 14

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Considering $\vec{u}, \vec{v}, \vec{w} \in V^3 \lambda \in \mathbb{R}$, then

\Vert \vec{u} \cdot \vec{v}\Vert \leq  \Vert\vec{u}\Vert  \Vert\vec{v}\Vert$ we have $(\vec{u} \cdot \vec{v})^2 \leq (\vec{u} \cdot \vec{u})(\vec{v} \cdot \vec{v}) \quad$

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We have the equation

\dfrac{\sin ^4 x }{a} + \dfrac{\cos^4 x }{b}  = \dfrac{1}{a+b}, a,b\in\mathbb{N}

We can use the Cauchy–Schwarz  Inequality because a and b are greater than 0. In fact, a>0 \wedge b>0 \implies ab>0. Using the Cauchy–Schwarz  Inequality, we have

\dfrac{\sin ^4 x }{a} + \dfrac{\cos^4 x }{b}   =\dfrac{(\sin^2 x)^2}{a}+\dfrac{(\cos^2 x)}{b}\geq \dfrac{(\sin^2 x+\cos^2 x)^2}{a+b} = \dfrac{1}{a+b}

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\dfrac{\sin^2{x}}{a}=\dfrac{\cos^2{x}}{b}=\dfrac{1}{a+b}

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Therefore, once we can write

\sin^2 x = \dfrac{a}{a+b} \implies \sin^{4n}x = \dfrac{a^{2n}}{(a+b)^{2n}} \implies\dfrac{\sin^{4n}x }{a^{2n-1}} = \dfrac{a^{2n}}{(a+b)^{2n}\cdot a^{2n-1}}

It is the same thing for cosine, thus

\cos^2 x = \dfrac{b}{a+b} \implies \dfrac{\cos^{4n}x }{b^{2n-1}} = \dfrac{b^{2n}}{(a+b)^{2n}\cdot b^{2n-1}}

Once

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dividing both numerator and denominator by (a+b), we get

\dfrac{a+b}{(a+b)^{2n} } =  \dfrac{1}{(a+b)^{2n-1} }

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\dfrac{\sin ^{4n} x }{a^{2n-1}} + \dfrac{\cos^{4n} x }{b^{2n-1}}  = \dfrac{1}{(a+b)^{2n-1}}, a,b\in\mathbb{N}

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