Question

The function $f:[0,3]\to [1,29]$, defined by $f\left(x\right)=2x^3–15x^2+36x+1$, is

• A. Neither one-one nor onto
• B. One-one but not onto
• C. Onto but not one-one
• D. One-one and onto

Answer 1

The correct option is C Onto but not one-one

$f\left(x\right)=2x^3-15x^2+36x+1$
$f^{\prime }\left(x\right)=6x^2–30x+36$
$=6(x^2–5x+6)$
$=6(x–2)(x–3)$
Clearly the derivative changes sign in $[0,3]$ so, $f$ is NOT one-one.
Now the function is increasing in $[0,2]$ and decreasing in $[2,3]$
Also, $f\left(0\right)=1$
$f\left(2\right)=29$
$f\left(3\right)=8$
Hence, the range is $[1,29]$ and so, the function is onto.