QuestionProb. 5 The function $f(x)=x^{2}$ is not invertible, as it is possible for two different numbers to have the same square. (For instance, $2^{2}=4$ and $(-2)^{2}=4$ .) However, in spite of this, we still like to talk about the square root function $g(x)=\sqrt{x}$ . (a) What is $\sqrt{9}$ ? Is this the only number whose square is 9 ? (b) For which values of $x$ is it true that $\sqrt{x^{2}}=x$ ?

Studdy Solution

STEP 1

1. The function $f(x) = x^2$ is not one-to-one, meaning it does not have an inverse over all real numbers.
2. The square root function $g(x) = \sqrt{x}$ typically refers to the principal (non-negative) square root.
3. We need to consider both positive and negative roots when discussing squares.

STEP 2

1. Solve part (a) to find $\sqrt{9}$ and discuss other numbers whose square is 9.
2. Solve part (b) to determine for which values of $x$ it is true that $\sqrt{x^2} = x$ .

STEP 3

For part (a), calculate $\sqrt{9}$ :
The principal square root of 9 is 3, because:
$\sqrt{9} = 3$

STEP 4

Discuss if there are other numbers whose square is 9:
The number $-3$ also satisfies the equation because:
$(-3)^2 = 9$
Thus, both $3$ and $-3$ are numbers whose square is 9.

STEP 5

For part (b), determine for which values of $x$ it is true that $\sqrt{x^2} = x$ :
The expression $\sqrt{x^2}$ gives the principal (non-negative) square root, which means:
$\sqrt{x^2} = |x|$
For $\sqrt{x^2} = x$ to hold true, $x$ must be non-negative. Therefore, the condition is:
$x \geq 0$
The solution to the problem is: (a) $\sqrt{9} = 3$ , and the numbers whose square is 9 are $3$ and $-3$ . (b) $\sqrt{x^2} = x$ is true for $x \geq 0$ .

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