What is the range of f(x)=3sqrt(-2x^2+72) ?

The range of f(x)=3sqrt(-2x^2+72) is 0<= f(x)<= 18sqrt(2)

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range of f(x)=3sqrt(-2x^2+72)

range

f (x) = 3 - 2 x^{2} + 72

0 \leq f (x) \leq 182

Interval Notation

[0, 182]

Solution steps

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

Domain of

3 - 2 x^{2} + 72 : - 6 \leq x \leq 6

Extreme Points of

3 - 2 x^{2} + 72 :

Minimum

(- 6, 0),

Maximum

(0, 182),

Minimum

(6, 0)

Find the range for the interval

- 6 \leq x \leq 6 : 0 \leq f (x) \leq 182

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

0 \leq f (x) \leq 182

(1, - 7), (1, - 12)

f (x) = x^{2} - 14

f (x) = 7 x^{2} ln (\frac{x}{4})

f (x) = x^{2} - 6 x

f (x) = - 5 x^{2} + 2 x + 8

What is the range of f(x)=3sqrt(-2x^2+72) ?
The range of f(x)=3sqrt(-2x^2+72) is 0<= f(x)<= 18sqrt(2)