To evaluate the function [tex]\(h(r) = -8^r\)[/tex] when [tex]\(r = \frac{1}{3}\)[/tex], we proceed with the following steps:
1. Substitute the value of [tex]\( r \)[/tex] into the function:
[tex]\[
h\left(\frac{1}{3}\right) = -8^{\frac{1}{3}}
\][/tex]
2. Evaluate the expression [tex]\( -8^{\frac{1}{3}} \)[/tex]:
- First, we need to consider what [tex]\( 8^{\frac{1}{3}} \)[/tex] means. The exponent [tex]\(\frac{1}{3}\)[/tex] denotes the cube root of 8.
- The cube root of 8 is 2, since [tex]\( 2^3 = 8 \)[/tex].
- Therefore, [tex]\( 8^{\frac{1}{3}} = 2 \)[/tex].
3. Incorporate the negative sign:
- The given function is [tex]\( -8^{\frac{1}{3}} \)[/tex].
- This implies that we need the principal cube root of -8.
4. Consider the principal root of a negative number:
- The cube root of a negative number, like -8, can result in a real number, but we must be aware that operating directly can sometimes cause confusion with complex numbers.
5. Assess the result in the real number system:
- In the real number system, the cube root of -8 is often treated as undefined in standard contexts (as it may involve complex calculations).
Given these considerations, we conclude:
- The evaluation of [tex]\( h\left( \frac{1}{3} \right) \)[/tex] in the context provided shows that the solution is not a well-defined real number within simple algebraic operations as we are dealing with the root of a negative number.
Thus, the conclusion is:
[tex]\[
h\left( \frac{1}{3} \right) = \text{DNE} \ (\text{Does Not Exist})
\][/tex]
This is because the cube root of a negative number leads to results not well-defined in basic real arithmetic contexts provided here.
