What is the domain and range of $h (x) = - \sqrt{x^{2} - 16} - 3$ $h(x)=-sqrt(x^2-16) -3$ ?

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What is the domain and range of $h (x) = - \sqrt{x^{2} - 16} - 3$ $h(x)=-sqrt(x^2-16) -3$ ?

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Answer 1 · 2017-12-29T14:53:09

Domain for $h (x)$ $h(x)$ is $x \leq - 4$ $x<=-4$ and $x \geq 4$ $x>=4$ . Range for $h (x)$ $h(x)$ is $(- \infty, - 3)$ $(-oo,-3)$ .

Explanation:

It is apparent that $x^{2} - 16 > 0$ $x^2-16>0$ , hence we must $x \leq - 4$ $x<=-4$ or $x \geq 4$ $x>=4$ and that is the domain for $h (x)$ $h(x)$ .

Further the least value for $\sqrt{x^{2} - 16}$ $sqrt(x^2-16)$ is $0$ $0$ and it can up to $\infty$ $oo$ .

Hence range for $h (x) = - \sqrt{x^{2} - 16} - 3$ $h(x)=-sqrt(x^2-16)-3$ is from a minimum of $- \infty$ $-oo$ to maximum of $- 3$ $-3$ i.e. $(- \infty, - 3)$ $(-oo,-3)$ .

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What is the domain and range of $h (x) = - \sqrt{x^{2} - 16} - 3$ $h(x)=-sqrt(x^2-16) -3$ ?

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What is the domain and range of h(x)=-sqrt(x^2-16) -3h(x)=−√x2−16−3h(x)=-sqrt(x^2-16) -3?

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What is the domain and range of $h (x) = - \sqrt{x^{2} - 16} - 3$ $h(x)=-sqrt(x^2-16) -3$ ?