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Given that,

- $f(x) = x^{2}-7x$
- $g(x) = x+3$

Now, $f(g(x))-3x = f(x+3)-3x = (x+3)^{2} – 7(x+3)-3x$

$\qquad \qquad \qquad = x^{2}+9+6x-7x-21-3x = x^{2}-4x-12$

Let $h(x) = x^{2}-4x-12$

$\Rightarrow h’(x) = 2x-4$

For minimum value of $x:$

$\Rightarrow h’(x) = 0.$

$\Rightarrow 2x-4=0$

$\Rightarrow \boxed{x=2}$

Now, $h(2) = 2^{2}-4(2)-12$

$\Rightarrow h(2)=4-8-12.$

$\Rightarrow \boxed{h(2) = -16}$

$\therefore$ The minimum possible value of $f(g(x))-3x$ is $-16.$

Correct Answer $:\text{A}$