Consider the following functions from R to R, where R is the set of real numbers:
(A) \(f(x) = 7 - 3x\)
(B) \(f(x) = x^{2} + 4\)
(C) \(f(x) = e^{x}\)
(D) \(f(x) = x^{3} + 1\)
Which of the above functions are onto?
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (D) only
A function from R to R is onto when every real number y is an image, that is, when \(f(x) = y\) has a real solution for every real y.
For \(7 - 3x\):
\( \displaystyle x = \frac{7 - y}{3} \)
For \(x^{3} + 1\):
\(x = \sqrt[3]{y - 1}\), a real number for every real y
For \(x^{2} + 4\): \(x^{2} \ge 0\), so every image is at least 4. For \(e^{x}\): every image is positive.
So \(7 - 3x\) and \(x^{3} + 1\) are onto, while the range of \(x^{2} + 4\) is \([4, \infty)\) and the range of \(e^{x}\) is \((0, \infty)\).
Checking the statements
(A) \(f(x) = 7 - 3x\) — Correct. \(x = \frac{7 - y}{3}\) gives \(f(x) = y\).
(B) \(f(x) = x^{2} + 4\) — Incorrect. No real x gives \(f(x) = 0\), since \(f(x) \ge 4\).
(C) \(f(x) = e^{x}\) — Incorrect. No real x gives \(e^{x} = -1\).
(D) \(f(x) = x^{3} + 1\) — Correct. \(x = \sqrt[3]{y - 1}\) gives \(f(x) = y\).