How do you find the global extreme values for # y=x^2-6x-1# on [-2,2]?

Question

Emilia Aguilar · Accepted Answer

#f_min (2)=-9#

#f_max (-2)=15# The possible extreme points of such a function are: 1) vertex of the parabola #V(p,q)# 2) end points of the interval (in this case #a=-2, b=2#) First we calculate the vertex: #p=-b/(2a)=6/2=3# 3 does not belong to the interval mentioned in the task, so we don't have to calculate #q#, we know, that the extreme values are at the ends of the interval. So we calculate: #f(-2)=4+12-1=15# #f(2)=4-12-1=-9# Now we can write the answer. #f_min (2)=-9# #f_max (-2)=15#

Evelyn Agard · Answer

undefined To find the global extreme values of $ y = x^2 - 6x - 1 $ on the interval $[-2, 2]$, we first find the critical points by taking the derivative of the function and setting it equal to zero. Then we evaluate the function at the critical points and the endpoints of the interval to determine the global maximum and minimum values.

1. Find the derivative of the function: $ y' = 2x - 6 $.
2. Set the derivative equal to zero and solve for $ x $: $ 2x - 6 = 0 $.
3. Solve for $ x $: $ x = 3 $.

Now we evaluate the function at the critical point and endpoints of the interval:

- Evaluate $ y $ at $ x = -2 $: $ y(-2) = (-2)^2 - 6(-2) - 1 = 4 + 12 - 1 = 15 $.
- Evaluate $ y $ at $ x = 2 $: $ y(2) = 2^2 - 6(2) - 1 = 4 - 12 - 1 = -9 $.
- Evaluate $ y $ at $ x = 3 $: $ y(3) = 3^2 - 6(3) - 1 = 9 - 18 - 1 = -10 $.

Comparing these values, the global maximum value is 15, which occurs at $ x = -2 $, and the global minimum value is -10, which occurs at $ x = 3 $.

Evelyn Agard · Answer

undefined To find the global extreme values of the function $y = x^2 - 6x - 1$ on the interval $[-2, 2]$, we follow these steps:

1. Find the critical points of the function within the given interval by setting the derivative equal to zero and solving for $x$.
2. Evaluate the function at these critical points as well as at the endpoints of the interval.
3. Compare the values obtained in step 2 to determine the global maximum and minimum.

First, let's find the derivative of the function:

$$ y = x^2 - 6x - 1 $$
$$ \frac{dy}{dx} = 2x - 6 $$

Now, set the derivative equal to zero to find critical points:

$$ 2x - 6 = 0 $$
$$ 2x = 6 $$
$$ x = 3 $$

The critical point is $x = 3$.

Now, we evaluate the function at the critical point and endpoints of the interval:

$$ y(-2) = (-2)^2 - 6(-2) - 1 = 4 + 12 - 1 = 15 $$
$$ y(2) = 2^2 - 6(2) - 1 = 4 - 12 - 1 = -9 $$
$$ y(3) = 3^2 - 6(3) - 1 = 9 - 18 - 1 = -10 $$

Thus, the function values are $y(-2) = 15$, $y(2) = -9$, and $y(3) = -10$.

The global maximum value is 15, occurring at $x = -2$, and the global minimum value is -10, occurring at $x = 3$.