Question

If f(x^3-6x^2+11x+8) =x^2-+1 then find the sum of all possible values of f(14)

Answer 1

Step-by-step explanation:

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the above answer is correct

Answer 2

The sum of all possible values of f(14) is 17.

Given:

$f(x^{3} -6x^{2} +11x+8)=x^{2} +1$

To Find:

We have to find the sum of all possible values of  $f(14)$.

Solution:

$x^{3} - 6x^{2} +11x+8$

$= (x^{3} -6x^{2} +11x-6)+14$

$={(x-1)(x^{2} -5x+6)}+14$

$={(x-1)(x^{2} -2x-3x+6)}+14$

$=(x-1)(x-2)(x-3)+14$ ....................(i)

∵$f(x^{3} -6x^{2}+11x+8)=x^{2} +1$

In order to find f(14), we have to find the value of x for which $x^{3}-6x^{2} +11x+8=14$.

From (i),

$f(x^{3}-6x^{2} +11x+8)=f((x-1)(x-2)(x-3)+14)=x^{2} +1$ .................(ii)

Putting x=1 in equation (ii), we get,

$f(14)=1^{2} +1 = 2$ .............(A)

Putting x=2 in the equation (ii), we get,

$f(14)=2^{2} +1=5$...............(B)

Putting x=3 in equation (ii), we get,

$f(14)=3^{2}+1 = 10$.............(C)

From (A), (B), and (C), the sum of all possible values of $f(14)=2+5+10=17$

Hence, the sum of all possible values of f(14) is 17.

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