Name: Antilogarithm
Uploaded: 2019-07-19
Duration: 4 min 44 s
Description: Antilogarithm is the inverse technique of finding the same number. Sounds confused?! This video will help you understand antilogarithm and how to solve probl...

Question

For

f : R \to [\frac{3}{4}, \infty)

, is

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

invertible? If not, in which region it is invertible?

Accepted Answer

Given: $y = f (x) = x^{2} + x + 1$

Discriminant $(D) = b^{2} - 4 a c$

Here $a = 1, b = 1, c = 1$

$\Rightarrow D = 1 - 4 = - 3$

Range of quadratic expression is $[\frac{- D}{4 a}, \infty)$ , for $a > 0$

$\frac{- D}{4 a} = - \frac{- (- 3)}{4 \cdot 1} = \frac{3}{4}$

Range of $f (x)$ is $[\frac{3}{4}, \infty)$

$\Rightarrow f (x) \geq 0, \forall x \in R$

Also, $x^{2} + x + 1$

$= {(x + \frac{1}{2})}^{2} + \frac{3}{4} \geq \frac{3}{4}$

Minimum value of $y = f (x)$ occurs at $x = - \frac{1}{2}$

There exists horizontal line which is cutting the graph of $y = f (x)$ at more than one point so it's not a one-one function.

So, function is not an invertible function, as for a function to be invertible it must be one-one and onto.

Case $(1) x \geq - \frac{1}{2},$

$\Rightarrow f : [- \frac{1}{2}, \infty) \to [\frac{3}{4}, \infty)$

$f$ is an increasing function, as $f^{'} (x) > 0$ in this interval, so function is one-one and also an onto function because range $=$ co-domain

Case $(2) x \leq - \frac{1}{2},$

$\Rightarrow f : (- \infty, - \frac{1}{2}] \to [\frac{3}{4}, \infty)$

$f$ is a decreasing function, as $f^{'} (x) < 0$ in this interval so, $y = f (x)$ is a one-one function and also an onto function because range $=$ co-domain

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