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# domain

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Find the domain of the function $$f(x)=\sqrt{-10x^2-11x+6+9x^2-5x}.$$

Aug 22, 2023

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To find the domain of the function $$f(x) = \sqrt{-10x^2-11x+6+9x^2-5x}$$, we need to determine the values of $$x$$ for which the expression under the square root is defined.

Simplify the expression under the square root:

$-10x^2 - 11x + 6 + 9x^2 - 5x = -x^2 - 16x + 6.$

The expression under the square root must not be negative, as the square root of a negative number is not a real number. Therefore, we need to find the values of $$x$$ for which $$-x^2 - 16x + 6 \geq 0$$.

To solve this inequality, we can factor the quadratic expression:

$-x^2 - 16x + 6 = -(x^2 + 16x - 6).$

Now, we want to find the values of $$x$$ that make the quadratic expression $$x^2 + 16x - 6$$ non-negative. We can do this by finding the roots of the quadratic equation $$x^2 + 16x - 6 = 0$$ and determining the intervals where the expression is positive or zero.

Factoring the quadratic equation $$x^2 + 16x - 6 = 0$$ is a bit tricky, so we can use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$

In this case, $$a = 1$$, $$b = 16$$, and $$c = -6$$, so the solutions are:

$x = \frac{-16 \pm \sqrt{16^2 - 4 \cdot 1 \cdot (-6)}}{2 \cdot 1}.$

Simplifying this gives:

$x = -8 \pm \sqrt{130}.$

Since the quadratic expression $$x^2 + 16x - 6$$ opens upwards (the coefficient of $$x^2$$ is positive), it is non-negative in the interval between its roots. Therefore, the values of $$x$$ that satisfy $$-x^2 - 16x + 6 \geq 0$$ are given by:

$-8 - \sqrt{130} \leq x \leq -8 + \sqrt{130}.$

In interval notation, the domain of the function $$f(x)$$ is:

$$\boxed{[-8 - \sqrt{130}, -8 + \sqrt{130}]}.$$

Aug 23, 2023