What is the range of f(x)=ln(2x^2+x+1) ?

The range of f(x)=ln(2x^2+x+1) is f(x)>= ln(7/8)

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range of f(x)=ln(2x^2+x+1)

range

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

f (x) \geq ln (\frac{7}{8})

Interval Notation

[ln (\frac{7}{8}), \infty)

Solution steps

The function range is the combined domain of the inverse functions

Find the inverse function(s) of:

ln (2 x^{2} + x + 1)

Inverse of

ln (2 x^{2} + x + 1) : \frac{- 1 + 8 e ^{x} - 7}{4}, \frac{- 1 - 8 e ^{x} - 7}{4}

\frac{- 1 + 8 e ^{x} - 7}{4}, \frac{- 1 - 8 e ^{x} - 7}{4}

Find the domain of each inverse function

Domain of

\frac{- 1 + 8 e ^{x} - 7}{4} : x \geq ln (\frac{7}{8})

Domain of

\frac{- 1 - 8 e ^{x} - 7}{4} : x \geq ln (\frac{7}{8})

Combine the intervals

f (x) \geq ln (\frac{7}{8})

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

\frac{x - 5}{2 x + 4}

f (x) = \frac{3}{25 - x ^{2}}

f (x) = x^{2} 4

f (x) = \frac{x ^{2} + 5 x + 4}{- 2 x ^{2} - 6 x}

What is the range of f(x)=ln(2x^2+x+1) ?
The range of f(x)=ln(2x^2+x+1) is f(x)>= ln(7/8)