For real $x$ let , then$f(x) = {x^3} + 5x + 1$
A) f is one-one and onto R
B) f is neither one-one nor onto R
C) f is one-one but not onto R
D) f is onto R but not one-one
Answer
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Hint:The function is one-one if the function is strictly increasing. Differentiate the function and check for increasing and decreasing. A function is onto when for every element in the codomain there is at least one element in the domain.
Complete step-by-step answer:
We are given a function $f(x) = {x^3} + 5x + 1$
Here function is defined from R to R that is $f:R \to R$
We have to check whether the function is one-one and onto or not.
First, we check for onto.
Let
$
f(x) = {x^3} + 5x + 1 = y \\
f(x) = {x^3} + 5x + 1 - y = 0 \\
$
Now, we see that we have a cubic polynomial in $x$. We know that a polynomial of odd degree has at least one real root to any $y$ that belongs to codomain, then some $x$ which belong to the domain such that $f(x) = y$
Hence, function is onto.
Now, we check for one-one.
We know that if the function is strictly increasing then it is a one-one function.
We have $f(x) = {x^3} + 5x + 1$
If $f'(x) > 0$ then function is strictly increasing.
Differentiate the function with respect to $x$.
Use $\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}$
$f'(x) = 3{x^2} + 5$
Since, square of any number is always positive therefore, $f'(x) > 0$
It means the function is increasing.
Therefore, $f(x)$is one-one.
Hence, $f(x)$is one-one and onto R.
Option (A) is correct.
Note:We can also use the horizontal line test to check if the function is one-one or not.
The horizontal line test states that if a line is drawn parallel to the x-axis and it cuts the graph exactly at one point then the function is one-one. If the line cuts exactly at one point it means for every y-value in the function, there is a unique x-value.
Complete step-by-step answer:
We are given a function $f(x) = {x^3} + 5x + 1$
Here function is defined from R to R that is $f:R \to R$
We have to check whether the function is one-one and onto or not.
First, we check for onto.
Let
$
f(x) = {x^3} + 5x + 1 = y \\
f(x) = {x^3} + 5x + 1 - y = 0 \\
$
Now, we see that we have a cubic polynomial in $x$. We know that a polynomial of odd degree has at least one real root to any $y$ that belongs to codomain, then some $x$ which belong to the domain such that $f(x) = y$
Hence, function is onto.
Now, we check for one-one.
We know that if the function is strictly increasing then it is a one-one function.
We have $f(x) = {x^3} + 5x + 1$
If $f'(x) > 0$ then function is strictly increasing.
Differentiate the function with respect to $x$.
Use $\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}$
$f'(x) = 3{x^2} + 5$
Since, square of any number is always positive therefore, $f'(x) > 0$
It means the function is increasing.
Therefore, $f(x)$is one-one.
Hence, $f(x)$is one-one and onto R.
Option (A) is correct.
Note:We can also use the horizontal line test to check if the function is one-one or not.
The horizontal line test states that if a line is drawn parallel to the x-axis and it cuts the graph exactly at one point then the function is one-one. If the line cuts exactly at one point it means for every y-value in the function, there is a unique x-value.
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