f(x) = square root of (x^2 - 10x - 11)find the largest possible domain

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neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To find the largewst possible domain of f(x) = sqrt(x^2-10x-11)

The domain of f(x) = sqrt(x^2-10x-11) is the set of all possible values for which f(x) is real.

The right sqrt (x^2-10x-11) is not real when (x^2-10x-11) < 0 (or negative).

sqrt(x^2-10x-11) real only when (x^2-10x-11) > 0.

Now we factorise x^2-10x-11.

x^2-10x-11= x^2-11x +x-11

x^2-11x+x -11 = x(x-11)+1(x-11)

(x-11)(x+1) .

Therefore x^-10x-11 = (x+1)(x-11) could be > 0 only when both factors (x+1) and (x-11) are negative or both factors (x+1) and (x-11)   are posive.

So this is possible only when x < -1  or x > 11.

Therefore the largest domain of x for f(x) to be real is x should belong to the interval  or domain (-infinity , -1) U (11 , +infinity).

william1941's profile pic

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

The domain of a function is defined as the values of x for which the function takes real values.

Here the function we have is g(x) = sqrt [ x^2 - 10x -11]

Now for g(x) to have a real value x^2 - 10x -11 has to be positive or equal to zero.

Therefore we can have all values of x except those for which x^2 - 10x -11 is negative or x^2 - 10x -11 < 0

Now, x^2 - 10x -11 < 0

=> x^2 - 11x + x -11 < 0

=> x(x-11) +1 (x-11) < 0

=> (x+1)(x-11) < 0

This is possible when only either one of (x+1) and (x-11) is less than zero.

For the case where x +1 < 0 and x-11 >0

=> x < -1 and x >11, this is not valid as x cannot be less than -1 and greater than 11 at the same time

For the case where x +1 >0 and x-11 < 0

=> x > -1 and x < 11

This is possible.

Therefore the domain of the function is all values of x except those where x > -1 and x < 11. Or the domain is all values of x except those that lie between -1 and 11.

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