(a) Find the critical values of the function.

Question

4. Given{ f(x) = (10 −1/4x^2 ,if x < 0

{5e^(x−2)/2 − x ,if x ≥ 0.

(a) Find the critical values of the function.

(b) Find the absolute extrema of f on the interval [−5, 5].

(c) Plot f on the interval [−5, 5].

Answer 1

a) Critical values are where the derivative equals zero and the slope changes positive to negative or negative to positive. For this one we've got two derivatives we'll be taking.

f'(x) = {-0.5x, x<0; and 2.5e^(x-2)/2 - 1; x>0

Note that there is no derivative at x=0 because the function is discontinuous at this point. For x<0, the derivative will never equal 0. For x>0, e^(x-2)/2 = 0.4, so x=2+2ln(0.4) ≈ 0.16742. For x between 0 and this value, the derivative is negative. For any value greater than it, the derivative is positive. This means that this is the only critical point for this function: (0.16742, 1.83258)

b) Absolute extrema will be at two of the following points: endpoints, discontinuities, and critical values. These would be x = -5, 0, 0.16742 (above), or 5. Simply plug these into the graph and determine whether it's the min/max value. Also, bear in mind that any of these values needs to be greater than 10 for for it to be the absolute max (we approach f(x)=10 as x goes to 0 from the left, but we never reach it).

x | f(x)

-5 | 3.75

0 | 1.84

0.16742 | 1.8325

5 | 17.40845

This means the absolute minimum occurs at the critical point from part (a) and the absolute maximum occurs at x=5 (f(5) = 17.40845).

c) The graph of this should be in the following parts.

[-5, 0) ⇒ The graph here is a fairly wide parabola opening downwards. The vertex at (0, 10) will be an open circle because the graph doesn't quite reach that point.

[0, 0.16742] ⇒ The graph goes down ever so slightly here before leveling off at the abs. min.

[0.16742, 5] ⇒ Starting from the abs. min., the graph grows exponentially greater until it reaches the abs. max. value.