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

v^{2} - 6 v + 8

\frac{x}{- 20} = 4.5

(\frac{2}{x ^{3}})^{2}

4 x^{2} + 2 x - 12

5 a^{2} - 25 a

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