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3 years ago

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

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

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

It is the same thing for cosine, thus

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