If the domain of the function f(x)= _e(4x^2+11x+6)+ ^-1(4x+3)+ ^-1((10x+6)/(3)) is ( , ], …

Question

If the domain of the function

$f(x)={\log }_{e}(4{x}^{2}+11x+6)+{\sin }^{-1}(4x+3)+{\cos }^{-1}(\frac{10x+6}{3})$ is $(\alpha ,\beta ]$, then

$36|\alpha +\beta |$ is equal to :

Accepted Answer

To find the domain of the function, we need to consider the individual functions and their respective domains. We have:

${f}_{1}(x)=\ln (4{x}^{2}+11x+6)$
${f}_{2}(x)={\sin }^{-1}(4x+3)$
${f}_{3}(x)={\cos }^{-1}(\frac{10x+6}{3})$

For ${f}_{1}(x)$:

$4{x}^{2}+11x+6>0$

Factoring the quadratic expression:

$(4x+3)(x+2)>0$

From this inequality, we have:

$x\in (-\infty,-2)\cup (-\frac{3}{4},\infty)$

For ${f}_{2}(x)$:

$-1\le 4x+3\le 1$

From these inequalities, we get:

$x\in [-1,-\frac{1}{2}]$

For ${f}_{3}(x)$:

$-1\le \frac{10x+6}{3}\le 1$

From these inequalities, we get:

$x\in [-\frac{9}{10},-\frac{3}{10}]$

Now, we need to find the intersection of the domains of the three functions:

$(-\infty,-2)\cup (-\frac{3}{4},\infty)\cap [-1,-\frac{1}{2}]\cap [-\frac{9}{10},-\frac{3}{10}]$

To find the intersection, let's analyze the intervals:

The interval $(-\infty,-2)\cup (-\frac{3}{4},\infty)$ contains all $x$ values less than $-2$ and greater than $-\frac{3}{4}$.
The interval $[-1,-\frac{1}{2}]$ contains all $x$ values between $-1$ and $-\frac{1}{2}$.
The interval $[-\frac{9}{10},-\frac{3}{10}]$ contains all $x$ values between $-\frac{9}{10}$ and $-\frac{3}{10}$.

Looking at the intervals, we can see that the intersection is:

$x\in (-\frac{3}{4},-\frac{1}{2}]$

Thus, the domain of the function is $(\alpha ,\beta ]=(-\frac{3}{4},-\frac{1}{2}]$. Now, we need to find the value of $36|\alpha +\beta |$:

$36|-\frac{3}{4}-\frac{1}{2}|=36|-\frac{5}{4}|=45$

Solutions · 1