What is 1<= ((3x^2-7x+8))/((x^2+1))<= 2 ?

The solution to 1<= ((3x^2-7x+8))/((x^2+1))<= 2 is 1<= x<= 6

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1<= ((3x^2-7x+8))/((x^2+1))<= 2

1 \leq \frac{( 3 x ^{2} - 7 x + 8 )}{( x ^{2} + 1 )} \leq 2

1 \leq x \leq 6

Interval Notation

[1, 6]

Solution steps

1 \leq \frac{( 3 x ^{2} - 7 x + 8 )}{( x ^{2} + 1 )} \leq 2

a \leq u \leq b

then

a \leq u and u \leq b

1 \leq \frac{( 3 x ^{2} - 7 x + 8 )}{( x ^{2} + 1 )} and \frac{( 3 x ^{2} - 7 x + 8 )}{( x ^{2} + 1 )} \leq 2

1 \leq \frac{3 x ^{2} - 7 x + 8}{x ^{2} + 1} :

True for all

x

\frac{3 x ^{2} - 7 x + 8}{x ^{2} + 1} \leq 2 : 1 \leq x \leq 6

Combine the intervals

True for all x and 1 \leq x \leq 6

Merge Overlapping Intervals

1 \leq x \leq 6

2 < ∣ x ∣ \leq 5

- x < x^{2} < 6

2 x - 3 < 3 x - 5 < 2 x + 3

- 2 \leq \frac{1}{2} - 2 \frac{x}{3} \leq \frac{11}{6}

\frac{( - x + 3 )}{2 - 2} \leq \frac{( 3 ( x + 2 ) )}{5} \leq \frac{x}{6}

What is 1<= ((3x^2-7x+8))/((x^2+1))<= 2 ?
The solution to 1<= ((3x^2-7x+8))/((x^2+1))<= 2 is 1<= x<= 6