What are the absolute extrema of # f(x)= 2 + x^2 in [-2, 3]#?

Question

Christopher Allers · Accepted Answer

#f(x)# has an absolute minimum of 2 at #x=0#

#f(x)= 2+x^2#

#f(x)# is a parabola with a single absolute minimum where #f'(x)=0#

#f'(x) =0+2x = 0 -> x=0#

#:.f_min(x) = f(0) = 2#

This can be seen on the graph of #f(x)# below:

graph{2+x^2 [-9.19, 8.59, -0.97, 7.926]}

Charles Arocha · Answer

undefined To find the absolute extrema of $ f(x) = 2 + x^2 $ in the interval $[-2, 3]$, we first need to evaluate the function at the endpoints of the interval and at any critical points within the interval.

1. Evaluate $ f(x) $ at the endpoints:
   $ f(-2) = 2 + (-2)^2 = 2 + 4 = 6 $
   $ f(3) = 2 + (3)^2 = 2 + 9 = 11 $

2. Find critical points by taking the derivative of $ f(x) $ and setting it equal to zero:
   $ f'(x) = 2x $
   Setting $ f'(x) = 0 $ gives $ x = 0 $.

3. Evaluate $ f(x) $ at the critical point:
   $ f(0) = 2 + (0)^2 = 2 $

Comparing the values, we find that the absolute minimum is $ f(-2) = 6 $ and the absolute maximum is $ f(3) = 11 $.

Elijah Abram · Answer

undefined To find the absolute extrema of $ f(x) = 2 + x^2 $ in the interval $[-2, 3]$, we need to evaluate the function at the endpoints of the interval as well as at any critical points within the interval.

The critical points occur where the derivative of the function is equal to zero or undefined.

The derivative of $ f(x) $ is $ f'(x) = 2x $.

Setting $ f'(x) $ equal to zero, we find the critical point to be $ x = 0 $.

Next, we evaluate $ f(x) $ at the endpoints and the critical point:

- At $ x = -2 $, $ f(-2) = 2 + (-2)^2 = 6 $
- At $ x = 0 $, $ f(0) = 2 + 0^2 = 2 $
- At $ x = 3 $, $ f(3) = 2 + 3^2 = 11 $

Comparing the values, the absolute maximum occurs at $ x = 3 $ where $ f(x) = 11 $, and the absolute minimum occurs at $ x = 0 $ where $ f(x) = 2 $.