Score: 0 of 3 points Given function \( f(x)=8 x^{p}-6 \), which integer values of \( p \) in range \( [2,6] \) make the function one-to-one. Answer all values, separated by commas. Example: \( 1,2,3 \).

To determine the integer values of \( p \) in the range \( [2,6] \) that make the function \( f(x) = 8x^p - 6 \) one-to-one, we need to analyze the behavior of the function based on the exponent \( p \). A function is one-to-one if it is either strictly increasing or strictly decreasing. This is typically guaranteed if the derivative does not change sign. The derivative of \( f(x) \) is given by: \[ f'(x) = 8p x^{p-1} \] - For \( p = 2 \): \( f'(x) = 16x \), which is zero at \( x = 0 \), thus not one-to-one (increasing on \( x > 0 \) and decreasing on \( x < 0 \)). - For \( p = 3 \): \( f'(x) = 24x^2 \), which is always non-negative and only equals zero at \( x = 0 \). This still fails the one-to-one test when considering the entire set of real numbers. - For \( p = 4 \): \( f'(x) = 32x^3 \), which has the same issue as \( p = 2 \) (horizontal inflection at \( x = 0 \)). - For \( p = 5 \): \( f'(x) = 40x^4 \), like \( p = 3 \) and \( p = 4 \), this becomes non-negative and isn’t one-to-one across all \( x \). - For \( p = 6 \): \( f'(x) = 48x^5 \), showing the same behavior as previous odd exponents. From the analysis, the function is one-to-one only when \( p \) is in the form of an odd integer \( > 1 \). Thus: The integer value of \( p \) in the range \( [2, 6] \) that makes the function one-to-one is \( 3, 5 \). Final Answer: \( 3, 5 \)