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2 years ago

F(x)=x^2+x-2 and g(x)=3x-1 Find (f+g)(1)=

Mathematics

1 answer:

Dimas [21]2 years ago

7 0

$( {x}^{2} + x - 2) + (3x - 1) \\ {x}^{2} + 4x - 3 \\ {1}^{2} + 4(1) - 3 \\ 1 + 4 - 3 \\ 2$

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xxTIMURxx [149]

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3 years ago

Order of operations what should be done first to evaluate 6+ (-5-7) ÷ 2 -8(3)

Maru [420]

Answer:-24

Step-by-step explanation:

6+(-5-7) ➗ 2-8(3)

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0-24=-24

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3 years ago

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3 years ago

WILL VOTE BRAINLIEST TO THE FIRST CORRECT ANSWER

KonstantinChe [14]

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1 year ago

Find the absolute extrema of f(x) = e^{x^2+2x}f ( x ) = e x 2 + 2 x on the interval [-2,2][ − 2 , 2 ] first and then use the com

fredd [130]

$f(x)=e^{x^2+2x}\implies f'(x)=2(x+1)e^{x^2+2x}$

$f$ has critical points where the derivative is 0:

$2(x+1)e^{x^2+2x}=0\implies x+1=0\implies x=-1$

The second derivative is

$f''(x)=2e^{x^2+2x}+4(x+1)^2e^{x^2+2x}=2(2x^2+4x+3)e^{x^2+2x}$

and $f''(-1)=\frac2e>0$ , which indicates a local minimum at $x=-1$ with a value of $f(-1)=\frac1e$ .

At the endpoints of [-2, 2], we have $f(-2)=1$ and $f(2)=e^8$ , so that $f$ has an absolute minimum of $\frac1e$ and an absolute maximum of $e^8$ on [-2, 2].

So we have

$\dfrac1e\le f(x)\le e^8$

$\implies\displaystyle\int_{-2}^2\frac{\mathrm dx}e\le\int_{-2}^2f(x)\,\mathrm dx\le\int_{-2}^2e^8\,\mathrm dx$

$\implies\boxed{\displaystyle\frac4e\le\int_{-2}^2f(x)\,\mathrm dx\le4e^8}$

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3 years ago