What is the minimum value of #g(x) = x^2-2x - 11/x?# on the interval #[1,7]#?

Question

What is the minimum value of  #g(x) = x^2-2x - 11/x?# on the interval #[1,7]#?

William Abdalla · Accepted Answer

The function is continuously increasing in the interval #[1,7]# its minimum value is at #x=1#.

It is obvious that #x^2-2x-11/x# is not defined at #x=0#, however it is defined in the interval #[1,7]#. Now derivative of #x^2-2x-11/x# is #2x-2-(-11/x^2)# or #2x-2+11/x^2# and it is positive throughout #[1,7]# Hence, the function is continuously increasing in the interval #[1,7]# and as such minimum value of #x^2-2x-11/x# in the interval #[1,7]# is at #x=1#. graph{x^2-2x-11/x [-40, 40, -20, 20]}

Evelyn Agard · Answer

undefined To find the minimum value of $ g(x) = x^2 - 2x - \frac{11}{x} $ on the interval $[1, 7]$, we can first find the critical points within this interval by taking the derivative of $ g(x) $, setting it equal to zero, and solving for $ x $. Then, we evaluate $ g(x) $ at these critical points as well as at the endpoints of the interval $[1, 7]$. The smallest value among these will be the minimum value of $ g(x) $ on the given interval.

1. Find the derivative of $ g(x) $:
$$ g'(x) = 2x - 2 + \frac{11}{x^2} $$

2. Set $ g'(x) $ equal to zero and solve for $ x $:
$$ 2x - 2 + \frac{11}{x^2} = 0 $$
$$ 2x^3 - 2x^2 + 11 = 0 $$

This equation may not have real roots. We can check if any exist within the interval $[1, 7]$.

3. Evaluate $ g(x) $ at critical points and endpoints:
   - $ g(1) = 1^2 - 2(1) - \frac{11}{1} = -12 $
   - $ g(7) = 7^2 - 2(7) - \frac{11}{7} $
   - Evaluate $ g(x) $ at any real roots found in step 2 if applicable.

The minimum value of $ g(x) $ on the interval $[1, 7]$ is the smallest value obtained from the evaluations above.