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If ${\text{f}}\left( {\text{x}} \right) = 3{x^3} - 9{x^2} - 27x + 15$ then find the maximum value of f(x).A)$- 66$ B)$30$ C)$- 30$ D)$66$

Last updated date: 19th Sep 2024
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Hint: First find the derivative of the function ${{\text{f}}^{'}}\left( {\text{x}} \right)$ and equate it to zero to find values of x. Then put these values in the second derivative ${{\text{f}}^{''}}\left( {\text{x}} \right)$ of the function. If ${{\text{f}}^{''}}\left( {\text{x}} \right)$$> 0 then f(x) has minima at x, If {{\text{f}}^{''}}\left( {\text{x}} \right)$$ < 0$ then f(x) has maxima at x.The maximum value value can be find by putting the value of maxima in f(x).

Given function, ${\text{f}}\left( {\text{x}} \right) = 3{x^3} - 9{x^2} - 27x + 15$-- (i)
On differentiating the given function w.r.t. x we get,
$\Rightarrow {{\text{f}}^{'}}\left( {\text{x}} \right) = \dfrac{d}{{dx}}\left( {3{x^3} - 9{x^2} - 27x + 15} \right)$
We know that $\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}$ and $\dfrac{{d\left( {{\text{constant}}} \right)}}{{dx}} = 0$
On applying this we get,
$\Rightarrow {{\text{f}}^{'}}\left( {\text{x}} \right) = 3 \times 3{x^{3 - 1}} - 9 \times 2{x^{2 - 1}} - 27{x^{1 - 1}} + 0$
On solving further we get,$\Rightarrow {{\text{f}}^{'}}\left( {\text{x}} \right) = 9{x^2} - 18x - 27$
On putting ${{\text{f}}^{'}}\left( {\text{x}} \right) = 0$ , we get-
$\Rightarrow 9{x^2} - 18x - 27 = 0$
On taking $9$ common and transferring on the right side we get-
$\Rightarrow {x^2} - 2x - 3 = 0$
On factoring we get,
$\Rightarrow {x^2} - 3x + x - 3 = 0$
$\Rightarrow \left( {{x^2} - 3x} \right) + \left( {x - 3} \right) = 0$
On taking x and 1 common we get,
$\Rightarrow x\left( {x - 3} \right) + 1\left( {x - 3} \right) = 0$
On taking $\left( {x - 3} \right)$ common we get,
$\Rightarrow \left( {x - 3} \right)\left( {x + 1} \right) = 0$
On putting both factors equal to zero we get,
$\Rightarrow x - 3 = 0 \Rightarrow x = 3$
or
$\Rightarrow x + 1 = 0 \Rightarrow x = - 1$
Now again differentiate the first derivative w.r.t. x-
$\Rightarrow {{\text{f}}^{''}}\left( {\text{x}} \right) = \dfrac{d}{{dx}}\left( {9{x^2} - 18x - 27} \right)$
On differentiating we get,
$\Rightarrow {{\text{f}}^{'}}\left( {\text{x}} \right) = 9 \times 2{x^{2 - 1}} - 18{x^{1 - 1}}$
On further solving we get,
$\Rightarrow {{\text{f}}^{''}}\left( {\text{x}} \right) = 18x - 18$ -- (ii)
Now put x=$3$ in eq. (ii)
$\Rightarrow {{\text{f}}^{''}}\left( 3 \right) = 18 \times 3 - 18$
On solving we get,
$\Rightarrow {{\text{f}}^{''}}\left( 3 \right) = 54 - 18 = 36$
Now here$\Rightarrow {{\text{f}}^{''}}\left( {\text{3}} \right) = 36 > 0$
So f(x) has minima at x= $3$
Now put x=$- 1$ in eq. (ii)
$\Rightarrow {{\text{f}}^{''}}\left( { - 1} \right) = 18\left( { - 1} \right) - 18$
On solving we get,
$\Rightarrow {{\text{f}}^{''}}\left( { - 1} \right) = - 18 - 18 = - 36$
Here,$\Rightarrow {{\text{f}}^{''}}\left( { - 1} \right) = - 36 < 0$
So f(x) has maxima at x=$- 1$
The maximum value of the function f(x) will be at x=$- 1$
So on putting x=$3$ in eq. (i) we get,
$\Rightarrow {\text{f}}\left( {\text{x}} \right) = 3{\left( { - 1} \right)^3} - 9{\left( { - 1} \right)^2} - 27\left( { - 1} \right) + 15$
On simplifying we get,
$\Rightarrow {\text{f}}\left( {\text{x}} \right) = - 3 - 9 + 27 + 15 \\ \Rightarrow {\text{f}}\left( {\text{x}} \right) = - 12 + 42 = 30 \\$
∴The function has maximum value $30$

Hence option B is the correct answer.

Note: Since f(x) has minima at x= $3$ so function also has minimum value. To find the minimum value put the value of x=$3$ in eq. (i)
$\Rightarrow {\text{f}}\left( x \right) = 3{\left( 3 \right)^3} - 9{\left( 3 \right)^2} - 27\left( 3 \right) + 15 \\ \Rightarrow {\text{f}}\left( x \right) = 3 \times 27 - 9 \times 9 - 27 \times 3 + 15 \\ \\$
On simplifying we get,
$\Rightarrow {\text{f}}\left( {\text{x}} \right) = 81 - 81 - 81 + 15 = - 81 + 15 = - 66$
So the minimum value of given function is $- 66$