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Math  /  Algebra

Question9. f(x)=11+x2f(x)=\frac{1}{1+x^{2}} бол f(−x),f(x−3),f(1x),2f(x),14f(x)+5f(1x)f(-x), f(x-3), f\left(\frac{1}{x}\right), 2 f(x), \frac{1}{4} f(x)+5 f\left(\frac{1}{x}\right)-ийг ол.

Studdy Solution

STEP 1

1. We are given the function f(x)=11+x2 f(x) = \frac{1}{1+x^2} .
2. We need to find expressions for f(−x) f(-x) , f(x−3) f(x-3) , f(1x) f\left(\frac{1}{x}\right) , 2f(x) 2f(x) , and 14f(x)+5f(1x) \frac{1}{4}f(x) + 5f\left(\frac{1}{x}\right) .

STEP 2

1. Substitute −x-x into f(x) f(x) to find f(−x) f(-x) .
2. Substitute x−3 x-3 into f(x) f(x) to find f(x−3) f(x-3) .
3. Substitute 1x \frac{1}{x} into f(x) f(x) to find f(1x) f\left(\frac{1}{x}\right) .
4. Multiply f(x) f(x) by 2 to find 2f(x) 2f(x) .
5. Combine 14f(x) \frac{1}{4}f(x) and 5f(1x) 5f\left(\frac{1}{x}\right) to find 14f(x)+5f(1x) \frac{1}{4}f(x) + 5f\left(\frac{1}{x}\right) .

STEP 3

To find f(−x) f(-x) , substitute −x -x into f(x) f(x) . f(−x)=11+(−x)2=11+x2 f(-x) = \frac{1}{1+(-x)^2} = \frac{1}{1+x^2}

STEP 4

To find f(x−3) f(x-3) , substitute x−3 x-3 into f(x) f(x) . f(x−3)=11+(x−3)2=11+(x2−6x+9)=1x2−6x+10 f(x-3) = \frac{1}{1+(x-3)^2} = \frac{1}{1+(x^2-6x+9)} = \frac{1}{x^2-6x+10}

STEP 5

To find f(1x) f\left(\frac{1}{x}\right) , substitute 1x \frac{1}{x} into f(x) f(x) . f(1x)=11+(1x)2=11+1x2=1x2+1x2=x2x2+1 f\left(\frac{1}{x}\right) = \frac{1}{1+\left(\frac{1}{x}\right)^2} = \frac{1}{1+\frac{1}{x^2}} = \frac{1}{\frac{x^2+1}{x^2}} = \frac{x^2}{x^2+1}

STEP 6

To find 2f(x) 2f(x) , multiply f(x) f(x) by 2. 2f(x)=2⋅11+x2=21+x2 2f(x) = 2 \cdot \frac{1}{1+x^2} = \frac{2}{1+x^2}

STEP 7

To find 14f(x)+5f(1x) \frac{1}{4}f(x) + 5f\left(\frac{1}{x}\right) , combine the expressions found in previous steps. 14f(x)=14⋅11+x2=14(1+x2) \frac{1}{4}f(x) = \frac{1}{4} \cdot \frac{1}{1+x^2} = \frac{1}{4(1+x^2)} 5f(1x)=5⋅x2x2+1=5x2x2+1 5f\left(\frac{1}{x}\right) = 5 \cdot \frac{x^2}{x^2+1} = \frac{5x^2}{x^2+1} 14f(x)+5f(1x)=14(1+x2)+5x2x2+1=14(1+x2)+5x2x2+1 \frac{1}{4}f(x) + 5f\left(\frac{1}{x}\right) = \frac{1}{4(1+x^2)} + \frac{5x^2}{x^2+1} = \frac{1}{4(1+x^2)} + \frac{5x^2}{x^2+1}
Solution:
1. f(−x)=11+x2 f(-x) = \frac{1}{1+x^2}
2. f(x−3)=1x2−6x+10 f(x-3) = \frac{1}{x^2-6x+10}
3. f(1x)=x2x2+1 f\left(\frac{1}{x}\right) = \frac{x^2}{x^2+1}
4. 2f(x)=21+x2 2f(x) = \frac{2}{1+x^2}
5. 14f(x)+5f(1x)=14(1+x2)+5x2x2+1 \frac{1}{4}f(x) + 5f\left(\frac{1}{x}\right) = \frac{1}{4(1+x^2)} + \frac{5x^2}{x^2+1}

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