A function `f:RrarrR` is defined by `f(x)=4x^(3)+5,x inR`.
Examine if f is one-one and onto.

Answer

To determine if the function \( f: \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = 4x^3 + 5 \) is one-one and onto, we will analyze each property step by step.

### Step 1: Check if the function is one-one

A function is one-one (injective) if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \).

1. Assume \( f(x_1) = f(x_2) \):
\[ 4x_1^3 + 5 = 4x_2^3 + 5 \]
2. Subtract 5 from both sides:
\[ 4x_1^3 = 4x_2^3 \]
3. Divide both sides by 4:
\[ x_1^3 = x_2^3 \]
4. Take the cube root of both sides:
\[ x_1 = x_2 \]

Since \( f(x_1) = f(x_2) \) leads to \( x_1 = x_2 \), the function \( f \) is one-one.

### Step 2: Check if the function is onto

A function is onto (surjective) if for every \( y \in \mathbb{R} \), there exists an \( x \in \mathbb{R} \) such that \( f(x) = y \).

1. Set \( f(x) = y \):
\[ y = 4x^3 + 5 \]
2. Rearranging gives:
\[ 4x^3 = y - 5 \]
3. Divide both sides by 4:
\[ x^3 = \frac{y - 5}{4} \]
4. Take the cube root:
\[ x = \sqrt[3]{\frac{y - 5}{4}} \]

Since \( y \) can be any real number, \( \frac{y - 5}{4} \) is also a real number, and taking the cube root of a real number yields another real number. Thus, for every \( y \in \mathbb{R} \), there exists an \( x \in \mathbb{R} \) such that \( f(x) = y \).

### Conclusion

Since \( f \) is both one-one and onto, we conclude that the function \( f(x) = 4x^3 + 5 \) is a bijection.

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