If the domain of the function f(x)=x^2-25(4-x^2)+ _10(x^2+2x-15) is (-, ) [ ,), then ^2+ ^…

Question

If the domain of the function

$f(x)=\frac{\sqrt{{x}^{2}-25}}{(4-{x}^{2})}+{\log }_{10}({x}^{2}+2x-15)$ is $(-\infty,\alpha )\cup [\beta ,\infty)$, then ${\alpha }^{2}+{\beta }^{3}$ is equal to :

Accepted Answer

To find the domain of the function 
$f(x)=\frac{\sqrt{{x}^{2}-25}}{(4-{x}^{2})}+{\log }_{10}({x}^{2}+2x-15),$ 
we need to consider the domain conditions for both the square root function and the logarithmic function.

The square root function $\sqrt{{x}^{2}-25}$ requires that the argument of the square root be non-negative, so
${x}^{2}-25\ge 0.$
This inequality is satisfied when
$x\le -5\, \, x\ge 5.$

The denominator of the rational part of $f(x)$, $(4-{x}^{2})$, cannot be zero, otherwise, the function will become undefined due to division by zero. Thus, we must have
$4-{x}^{2}
e 0.$
This inequality is violated when
$x=\pm 2.$

Combining these conditions gives us the domain for the rational part of the function:
$x\in (-\infty,-5]\cup (5,\infty)\, \, x
e 2,-2.$

Moving on to the logarithmic function, ${\log }_{10}({x}^{2}+2x-15)$, the argument must be positive:
${x}^{2}+2x-15>0.$
This is a quadratic inequality, which we can factor to find the solution:
$(x+5)(x-3)>0.$
From this, we see that the inequality is satisfied for
$x3.$

The overall domain of $f(x)$ is the intersection of the domains for each piece. Taking the intersection of the two sets gives us:
$x\in (-\infty,-5)\cup (5,\infty),$

Since the question states that the domain is of the form $(-\infty,\alpha )\cup [\beta ,\infty)$, we can infer that
$\alpha =-5\, \, \beta =5.$

We calculate ${\alpha }^{2}+{\beta }^{3}$ as follows:
${\alpha }^{2}+{\beta }^{3}=(-5{)}^{2}+{5}^{3}=25+125=150.$

So the correct answer, representing the sum of ${\alpha }^{2}$ and ${\beta }^{3}$, is:
Option D $150$.

Solutions · 1