What is the range of sqrt(6x^3+8x^2) ?

The range of sqrt(6x^3+8x^2) is f(x)>= 0

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range of sqrt(6x^3+8x^2)

range

6 x^{3} + 8 x^{2}

f (x) \geq 0

Interval Notation

[0, \infty)

Solution steps

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

Domain of

6 x^{3} + 8 x^{2} : x \geq - \frac{4}{3}

Extreme Points of

6 x^{3} + 8 x^{2} :

Minimum

(- \frac{4}{3}, 0),

Maximum

(- \frac{8}{9}, \frac{16 2}{9 3}),

Minimum

(0, 0)

Find the range for the interval

- \frac{4}{3} \leq x < \infty : 0 \leq f (x) < \infty

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

f (x) \geq 0

(- 1.2) (- 7)

f (x) = \frac{( 2 x ^{2} - 3 )}{( x ^{2} - 1 )}

f (x) = x^{2} - 2 x + 1

2 x^{3} - 5 x^{2} + 4 x + 2

x^{2} + x + 2