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The given function is \(f(x) = 9x^2 + 6x - 5\), defined for \(x \in R_{+}\), where \(R_{+}\) represents non-negative real numbers, and the codomain is \([-5, \infty)\).
To check if the function is one-one, we can analyze its derivative. The derivative of \(f(x)\) is \(f'(x) = 18x + 6\). Since \(x \in R_{+}\), \(x \geq 0\), so \(18x + 6 \geq 6 > 0\). This means that \(f'(x) > 0\) for all \(x \in R_{+}\). Therefore, the function is strictly increasing on \(R_{+}\). A strictly increasing function is always one-one (injective).
To check if the function is onto, we need to determine if the range of the function is equal to the codomain. The codomain is given as \([-5, \infty)\). Let's find the minimum value of the function. Since \(f(x) = 9x^2 + 6x - 5\), we can complete the square: \(f(x) = (3x)^2 + 2(3x)(1) + 1 - 1 - 5 = (3x + 1)^2 - 6\) Since \(x \in R_{+}\), the minimum value of \(3x + 1\) is 1 (when \(x = 0\)). Therefore, the minimum value of \((3x + 1)^2\) is \(1^2 = 1\). So, the minimum value of \(f(x)\) is \(1 - 6 = -5\). As \(x\) increases, \(f(x)\) also increases. Thus, the range of \(f(x)\) is \([-5, \infty)\), which is equal to the codomain. Therefore, the function is onto (surjective).
Since the function is both one-one (injective) and onto (surjective), it is bijective.
Final Answer: bijective