what the largest domain of the function,{(3-x)(2x+4)}^1/2?

Question

{(3-x)(2x+4)}^1/2

Answer 1

The domain of ((3-x)(2x+4))^1/2 is limited by the restriction on square roots (raising the quadratic function to the power of a half means square root) in that you can only take real roots (i.e., not imaginary) of positive quantities. Graphically the function is represented by an upturned U shape that dips below the x-axis (becomes negative) after intercepting the axis at x=-2 and 3. So these limits represent the domain. For example, x>3 makes 3-x negative and x<-2 makes 2x+4 negative. x must be between -2 and +3, including these limits.

Answer 2

Given {(3-x)(2x+4)}^1/2
Domain of the function
{x belongs to R:-2<=x<=3}

Hence the largest domain=3

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Answer 3

Given: {(3-x)(2x+4)}^½   This expression is also written in a radical form: √{(3-x)(2x+4)}

Domain,R, is a term that descibes the set of values, x∊R, for which a function of x, f(x) is defined.

Let f(x)=√{(3-x)(2x+4)} and g(x)=(3-x)(2x+4), so f(x)=√g(x).   Thus, the condition that defines the real values of f(x) is: g(x)≥0   We have: (3-x)(2x+4)≥0

The coefficient of x² in g(x) is negative(-2), so the graph of g(x) is convex upwards and intercepts x-axis(y=0) at x=-2 and x=3.   And (3-x)(2x+4)≥0, so the domain of x is: -2≤ x ≥3.

Therefore, the largest value of domain is: x=3