Question

If $f(x)=x^2−7x$ and $g(x)=x+3$, then the minimum value of $f(g(x))−3x$ is

• A. -20
• B. -15
• C. -12
• D. -16

Answer 1

Answer: (D)

-16

Answer 2

To find the minimum value of $f(g(x)) - 3x$, we first need to find the expression for $f(g(x))$.

Given:
$f(x) = x^2 - 7x$
$g(x) = x + 3$

Now, $f(g(x))$ is:
$f(g(x)) = (x+3)^2 - 7(x+3)$
$f(g(x)) = x^2 + 6x + 9 - 7x - 21$
$f(g(x)) = x^2 - x - 12$

Now, we are interested in:
$f(g(x)) - 3x = x^2 - x - 12 - 3x$
$f(g(x)) - 3x = x^2 - 4x - 12$

To find the minimum value of the above expression, we can find the derivative and set it to zero.
$\frac{d}{dx} (x^2 - 4x - 12) = 2x - 4$

Setting it to zero, $2x - 4 = 0$, gives $x = 2.$

Plugging x = 2 into $f(g(x)) - 3x$: $f(2+3) - 3(2) = 2^2 - 4(2) - 12 = 4 - 8 - 12 = -16$

So, the minimum value of $f(g(x)) - 3x$ is -16