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worty [1.4K]
3 years ago
11

The graph of the function f(x) = –(x 3)(x – 1) is shown below. which statement about the function is true? the function is posit

ive for all real values of x where x < –1. the function is negative for all real values of x where x < –3 and where x > 1. the function is positive for all real values of x where x > 0. the function is negative for all real values of x where x < –3 or x > –1.

Mathematics
2 answers:
sergiy2304 [10]3 years ago
9 0

Answer:

B

Step-by-step explanation:

Correct on e2020

kondaur [170]3 years ago
3 0
See attachment

positive is where it goes above x axis

so it is posittive for  x where -3<x<1
it is negativive where x<-3 and x>1



answer is
the function is negative for all real values of x where x < –3 and where x > 1


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solniwko [45]

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log x-\frac{log(x^{2}+1) }{2}-tan^{-1} x

Step-by-step explanation:

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[tex]\frac{1-x}{x(x^{2}+1) } =\frac{A(x^{2}+1)+(Bx+C)(x }{x(x^{2}+1) }......(1)

<u>step 2:-</u>

solving on both sides

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substitute x =0 value in equation (2)

1=A(1)+0

<u>A=1</u>

comparing x^2 co-efficient on both sides (in equation 2)

0 = A+B

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B=-1

comparing x co-efficient on both sides (in equation 2)

<u>-</u>1  =  C

<u>step 3:-</u>

substitute A,B,C values in equation (1)

now  

\\\int\limits^ {} \, \frac{1-x}{x(x^{2}+1) } d x =\int\limits^ {} \frac{1}{x} d x +\int\limits^ {} \frac{-x}{x^{2}+1 }  d x -\int\limits \frac{1}{x^{2}+1 }  d x

by using integration formulas

i)  by using \int\limits \frac{1}{x}   d x =log x+c........(a)\\\int\limits \frac{f^{1}(x) }{f(x)} d x= log(f(x)+c\\.....(b)

\int\limits tan^{-1}x  dx =\frac{1}{1+x^{2} } +C.....(c)

<u>step 4:-</u>

by using above integration formulas (a,b,and c)

we get answer is

log x-\frac{log(x^{2}+1) }{2}-tan^{-1} x

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