SOLUTION: For f(x) = SqRt(x^3 + 1) find f(-2)

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Question 957781: For f(x) = SqRt(x^3 + 1) find f(-2)
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29+=+sqrt%28x%5E3+%2B+1%29
f%28-2%29+=+sqrt%28%28-2%29%5E3+%2B+1%29
Simplifying...
f%28-2%29+=+sqrt%28-8+%2B+1%29
f%28-2%29+=+sqrt%28-7%29

The square root of -7 is not a real number. (IOW, it is impossible to square a real number and get -7). Since you posted this under the "real number" category then I assume you only want real number solutions. If so, the answer is: f(-2) is undefined. Or: -2 is not in the domain of f.

On the other hand, if you are looking for complex number solutions, then...
f%28-2%29+=+sqrt%28-1%2A7%29
f%28-2%29+=+sqrt%28-1%29%2Asqrt%287%29
f%28-2%29+=+i%2Asqrt%287%29
or
f%28-2%29+=+sqrt%287%29%2Ai
or, in a + bi form:
f%28-2%29+=+0+%2B+sqrt%287%29%2Ai

In response to the question in your thank you note: Algebra.com's website has a syntax which can be used Mathematical expressions. If you click on the "View Source" link just above this solution and you will see what I typed to get square roots displayed.