Show \(f(x)=e^x\) is One-One but Not Onto
📺 Video Explanation
📝 Question
Show that:
\[ f:\mathbb{R}\to\mathbb{R},\quad f(x)=e^x \]
is one-one but not onto.
Also discuss what happens if codomain is:
\[ \mathbb{R}_0^+=\{y\in\mathbb{R}:y>0\} \]
✅ Solution
🔹 Step 1: Prove One-One (Injective)
Assume:
\[ f(x_1)=f(x_2) \]
Then:
\[ e^{x_1}=e^{x_2} \]
Taking logarithm:
\[ x_1=x_2 \]
✔ Hence, \(f\) is one-one.
🔹 Step 2: Check Onto for Codomain \(\mathbb{R}\)
Since:
\[ e^x>0 \quad \text{for all } x\in\mathbb{R} \]
So function never gives:
- 0
- negative numbers
Thus range:
\[ (0,\infty) \]
But codomain is:
\[ \mathbb{R} \]
❌ Not onto.
🔹 Step 3: If Codomain is \(\mathbb{R}_0^+\)
Now codomain:
\[ (0,\infty) \]
For any:
\[ y>0 \]
choose:
\[ x=\ln y \]
Then:
\[ f(x)=e^{\ln y}=y \]
✔ Every positive real has pre-image.
So function becomes onto.
🎯 Final Answer
\[ \boxed{ f(x)=e^x:\mathbb{R}\to\mathbb{R} \text{ is one-one but not onto} } \]
If codomain is:
\[ (0,\infty) \]
then:
\[ \boxed{\text{f becomes bijective}} \]
🚀 Exam Shortcut
- Exponential function is strictly increasing
- Range of \(e^x\) is positive reals only
- Match codomain with range for bijection