What is the range of sqrt(6+x-2x^2) ?

The range of sqrt(6+x-2x^2) is 0<= f(x)<= 7/(2sqrt(2))

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range of sqrt(6+x-2x^2)

range

0 \leq f (x) \leq \frac{7}{2 2}

Interval Notation

[0, \frac{7}{2 2}]

Solution steps

Find the minimum and maximum value in each defined interval and unite the results

Domain of

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

Extreme Points of

6 + x - 2 x^{2} :

Minimum

(- \frac{3}{2}, 0),

Maximum

(\frac{1}{4}, \frac{7}{2 2}),

Minimum

(2, 0)

Find the range for the interval

- \frac{3}{2} \leq x \leq 2 : 0 \leq f (x) \leq \frac{7}{2 2}

Combine the ranges of all domain intervals to obtain the function range

0 \leq f (x) \leq \frac{7}{2 2}

What is the range of sqrt(6+x-2x^2) ?
The range of sqrt(6+x-2x^2) is 0<= f(x)<= 7/(2sqrt(2))