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Chapter 1: Problem 4

Evaluate (if possible) the function at the given value(s) of the independent variable. Simplify the results. \(f(x)=\sqrt{x+3}\) (a) \(f(-2)\) (b) \(f(6)\) (c) \(f(-5)\) (d) \(f(x+\Delta x)\)

Short Answer

Expert verified
The function evaluated at the given values is as follows: (a) \(f(-2) = 1\), (b) \(f(6) = 3\), (c) \(f(-5)\) is undefined, (d) \(f(x+\Delta x) = \sqrt{x+\Delta x+3}\).

Step by step solution

01

Evaluating Function at \(x=-2\)

First, replace 'x' in the function with the value -2 to obtain \(f(-2)=\sqrt{-2+3}\). This simplifies to \(f(-2)=\sqrt{1}\). The square root of 1 is 1, so \(f(-2) = 1\).
02

Evaluating Function at \(x=6\)

Next, replace 'x' in the function with the value 6 to obtain \(f(6)=\sqrt{6+3}\). This simplifies to \(f(6)=\sqrt{9}\). The square root of 9 is 3, so \(f(6) = 3\).
03

Evaluating Function at \(x=-5\)

For this value, replace 'x' in the function with the value -5 to obtain \(f(-5)=\sqrt{-5+3}\). This simplifies to \(f(-5)=\sqrt{-2}\). Because the square root of -2 is undefined (not a real number), we cannot evaluate this function at \(x=-5\).
04

Evaluating Function at \(x+\Delta x\)

Finally, replace 'x' in the function with \(x+\Delta x\) to obtain \(f(x+\Delta x)=\sqrt{x+\Delta x+3}\). This cannot be simplified further, so \(f(x+\Delta x)=\sqrt{x+\Delta x+3}\).

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