Prove \(f(x)=4x^3+7\) is a Bijection

📺 Video Explanation

📝 Question

Show that:

\[ f:\mathbb{R}\to\mathbb{R},\quad f(x)=4x^3+7 \]

is a bijection.

---

✅ Solution

🔹 Step 1: Prove One-One (Injective)

Assume:

\[ f(x_1)=f(x_2) \]

Then:

\[ 4x_1^3+7=4x_2^3+7 \]

Simplify:

\[ x_1^3=x_2^3 \]

Since cube function is strictly increasing:

\[ x_1=x_2 \]

✔ Hence, \(f\) is one-one.

---

🔹 Step 2: Prove Onto (Surjective)

Let:

\[ y\in\mathbb{R} \]

Need to find:

\[ x\in\mathbb{R} \]

such that:

\[ 4x^3+7=y \]

Solve:

\[ 4x^3=y-7 \]

\[ x^3=\frac{y-7}{4} \]

\[ x=\sqrt[3]{\frac{y-7}{4}} \]

Since cube root exists for every real number:

\[ x\in\mathbb{R} \]

✔ Hence, onto.

---

🎯 Final Answer

\[ \boxed{f(x)=4x^3+7\text{ is bijective}} \]

---

🚀 Exam Shortcut

• Odd degree cubic functions are strictly increasing
• Use inverse form to prove onto
• One-one + onto = bijection