Show that the function f : R → R : f (x) = x4 is neither one-one nor

1.	Show that the function f : R → R : f (x) = x4 is neither one-one nor onto.
Answer» To prove: function is neither one-one nor onto Given: f : R → R : f (x) = x⁴ We have, f(x) = x⁴ For, f(x₁) = f(x₂) ⇒ x₁⁴ = x₂⁴ ⇒ (x₁⁴ - x₂⁴) = 0 ⇒(x₁² - x₂²) (x₁² + x₂²) = 0 ⇒ (x₁ - x₂) (x₁ + x₂) (x₁² + x₂²) = 0 ⇒ x₁ = x₂ or, x₁ = -x₂ or, x₁² = -x₂² We are getting more than one value of x₁ (no unique image) ∴ f(x) is not one-one f(x) = x⁴ Let f(x) = y such that \(y\in R\) ⇒ y = x⁴ ^{\(\Rightarrow x=\sqrt[4]{y}\)} If y = -2, as \(y\in R\) Then x will be undefined as we can’t place the negative value under the square root Hence f(x) is not onto Hence Proved

Show that the function f : R → R : f (x) = x4 is neither one-one nor onto.

Answer»

To prove: function is neither one-one nor onto

Given: f : R → R : f (x) = x⁴

We have,

f(x) = x⁴

For, f(x₁) = f(x₂)

⇒ x₁⁴ = x₂⁴

⇒ (x₁⁴ - x₂⁴) = 0

⇒(x₁² - x₂²) (x₁² + x₂²) = 0

⇒ (x₁ - x₂) (x₁ + x₂) (x₁² + x₂²) = 0

⇒ x₁ = x₂ or, x₁ = -x₂ or, x₁² = -x₂²

We are getting more than one value of x₁ (no unique image)

∴ f(x) is not one-one

f(x) = x⁴

Let f(x) = y such that \(y\in R\)

⇒ y = x⁴

^{\(\Rightarrow x=\sqrt[4]{y}\)}

If y = -2, as \(y\in R\)

Then x will be undefined as we can’t place the negative value under the square root

Hence f(x) is not onto

Hence Proved