Question

The function $f:\left[ {0,3} \right] \to \left[ {1,29} \right]$, defined by $f\left( x \right) = 2{x^3} - 15{x^2} + 36x + 1$ is
A) One-one and onto
B) Onto but not one-one
C) One-one but not onto
D) Neither one-one nor onto

Answer 1

Hint: First, check whether the function is one-one or not. For that find the derivative of the function and set it equal to 0. Now check whether the function is strictly increasing or decreasing. After that check, the function is onto or not. For that find the range of the function and check whether the range is equal to the codomain or not.

Complete step-by-step answer:
Given: - $f:\left[ {0,3} \right] \to \left[ {1,29} \right]$
$f\left( x \right) = 2{x^3} - 15{x^2} + 36x + 1$
Check the function for one-one.
Find the first derivative of the function f(x),
$\begin{gathered}
  f'\left( x \right) = \dfrac{d}{{dx}}\left( {2{x^3} - 15{x^2} + 36x + 1} \right) \\
   = 6{x^2} - 30x + 36 \\
\end{gathered} $
Substitute the derivative equal to 0 to get the critical points,
$6{x^2} - 30x + 36 = 0$
Take out common factor from the left side,
$6\left( {{x^2} - 5x + 6} \right) = 0$
Divide both sides by 6,
${x^2} - 5x + 6 = 0$
Now, factor the equation to get the roots,
${x^2} - 2x - 3x + 6 = 0$
Take out common factors,
$x\left( {x - 2} \right) - 3\left( {x - 2} \right) = 0$
Take the common factors,
$\left( {x - 2} \right)\left( {x - 3} \right) = 0$
Equate $\left( {x - 2} \right)$ to 0,
$x - 2 = 0$
Move 2 to other sides,
$x = 2$
Now, equate $\left( {x - 3} \right)$ to 0,
$x - 3 = 0$
Move 3 to other sides,
$x = 3$
Thus, $x = 2,3$
$f\left( x \right)$ is increasing for $x < 2$ as $f'\left( x \right) > 0$ and $f\left( x \right)$ is decreasing for $2 < x < 3$ as $f'\left( x \right) < 0$.
Since $f\left( x \right)$ is not strictly increasing or strictly decreasing in the domain. So, $f\left( x \right)$ is not one-one.
Now, check the function for onto.
Since $f\left( x \right)$ is increasing in $\left[ {0,2} \right]$ and decreasing in $\left[ {2,3} \right]$.
Find $f\left( 0 \right)$,
$f\left( 0 \right) = 2{\left( 0 \right)^3} - 15{\left( 0 \right)^2} + 36\left( 0 \right) + 1$
Add and subtract the terms on the right side,
$\begin{gathered}
  f\left( 0 \right) = 0 + 0 + 0 + 1 \\
   = 1 \\
\end{gathered} $
Now, find $f\left( 2 \right)$,
$f\left( 2 \right) = 2{\left( 2 \right)^3} - 15{\left( 2 \right)^2} + 36\left( 2 \right) + 1$
Add and subtract the terms on the right side,
$\begin{gathered}
  f\left( 2 \right) = 16 - 60 + 72 + 1 \\
   = 29 \\
\end{gathered} $
Now, find $f\left( 3 \right)$,
$f\left( 3 \right) = 2{\left( 3 \right)^3} - 15{\left( 3 \right)^2} + 36\left( 3 \right) + 1$
Add and subtract the terms on the right side,
$\begin{gathered}
  f\left( 2 \right) = 54 - 135 + 108 + 1 \\
   = 28 \\
\end{gathered} $
Since the range of the function $f\left( x \right)$ $\left[ {1,29} \right]$ is equal to the codomain. So, $f\left( x \right)$ is onto.

Hence, option (B) is correct.

Note: The domain is the set of all possible x-values which will make the function "work", and will output real y-values.
The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually) after we have substituted the domain.
In a one-one function, given any y there is only one x that can be paired with the given y.
A function f from A to B is called onto if for all b in B there is a in A such that f(a) = b. That is, all elements in B are used.