What is the range of y=|2x^2+x-3| ?

The range of y=|2x^2+x-3| is f(x)>= 0

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range of y=|2x^2+x-3|

range

y = 2 x^{2} + x - 3

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

2 x^{2} + x - 3 : - \infty < x < \infty

Extreme Points of

2 x^{2} + x - 3 :

Minimum

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

Maximum

(- \frac{1}{4}, \frac{25}{8}),

Minimum

(1, 0)

Absolute value functions are piecewise functions

Find the function intervals:

x < - \frac{3}{2}, - \frac{3}{2} \leq x < 1, x \geq 1

Evaluate the function behavior for each interval

Find the range for the interval

- \infty < x < - \frac{3}{2} : 0 < f (x) < \infty

Find the range for the interval

- \frac{3}{2} \leq x < 1 : 0 \leq f (x) \leq \frac{25}{8}

Find the range for the interval

1 \leq x < \infty : 0 \leq f (x) < \infty

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

f (x) > 0 or 0 \leq f (x) \leq \frac{25}{8} or f (x) \geq 0

f (x) \geq 0

\frac{1}{4} cos (\frac{1}{4} x - \frac{π}{12})

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

f (x) = 36 - x^{2}

\frac{x + x ( 2 x ^{2} - 5 )}{2 x ^{2} - 5}

f (x) = 5 x^{\frac{1}{3}} + 4