Consider f : R_ + [ - 5, ) given by f(x) = 9x^ 2 + 6x - 5. Show t… - Vidyadip

Question

Consider $\text{f} : \text{R}_{+} \rightarrow [ - 5, \infty)$ given by $\text{f}(x) = 9x^{2} + 6x - 5.$ Show that f is invertible with $\text{f}^{-1}\text{(y)} = \bigg(\frac{\sqrt{\text{y} + 6} - 1}{3}\bigg).$
Hence Find:

$\text{f}^{-1} (10)$
$\text{y if }\text{f}^{-1} \text{(y)} = \frac{4}{3},$

where $R_+$ is the set of all non-negative real numbers.

✓

Answer

Clearly $\text{f}^{-1} \text{(y) = g (y):} [ -5, \infty) \rightarrow \text{R}_{+} \text{ and},$
$\text{fog (y) = y} \bigg(\frac{\sqrt{\text{y}\text{ + } 6} - 1}{3}\bigg) = 9 \bigg(\frac{\sqrt{\text{y + 6}} - 1}{3} \bigg)^{2} + 6 \bigg(\frac{\sqrt{\text{y + 6}} - 1}{3}\bigg) - 5 = \text{y}$
$\text{and (gof) (x) = g} (\text{9x}^{2} + \text{6x - 5)} = \frac{\sqrt{\text{9x}^{2} + \text{6x + 1}} - 1}{3} = \text{x}$
$\therefore \text{g = f}^{-1}$

$\text{f}^{-1} (10) = \frac{\sqrt{16} - 1}{3} = 1$
$\text{f}^{-1} (\text{y)} = \frac{4}{3} \Rightarrow \text{y = 19}$

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