0 reads0 ratings0 helpful
If (f:\mathbb{R}\to\mathbb{R}), (f(x)=x^2+e^x), why is (f) not onto?यदि (f:\mathbb{R}\to\mathbb{R}), (f(x)=x^2+e^x), तो (f) सर्वाच्छादक क्यों नहीं है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: Because (-1) is not an imageक्योंकि (-1) छवि नहीं बनता
Step 1: Since (x^2\ge0) and (e^x>0), we have (x^2+e^x>0). Step 2: The codomain (\mathbb{R}) contains (-1), but it cannot be an image of any (x). Step 3: Positivity or a lower bound can quickly disprove onto property.चरण 1: (x^2\ge0) और (e^x>0), इसलिए (x^2+e^x>0)। चरण 2: सहप्रांत (\mathbb{R}) में (-1) है, पर यह किसी भी (x) की छवि नहीं बन सकता। चरण 3: धनात्मकता या निचली सीमा दिखाकर सर्वाच्छादकता जल्दी नकारी जा सकती है।
Student feedback
Was this question useful?
Was this question useful?
👍 0 Helpful 👎 0 Not helpful
Difficulty
Easy0%
Medium0%
Hard0%
Was the explanation clear?
Yes 0% No 0%
0 responsesStudent Reviews
No published reviews yet.
