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

The Correct Option is D

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

Answer 2

We have 
\(f(g(x))−3x\)
\(⇒ f(x+3)-3x.\)
\(=  (x+3)^ 2 −7(x+3)−3x \)
\(=  x ^2 −4x−12\)
Value of expression: \(-\frac{D}{4a}\frac{ (4ac−b ^2)}{4a}\)
\(=- \frac{(16+48)}{4}\)
\(=-16\)