Q. 1 Global extrema on an interval: T... [FREE SOLUTION]

Chapter 3: Q. 1 (page 277)

Global extrema on an interval: The first-derivative test can be used to show that the function $f (x) = x^{2} + 3 x$ has a local minimum at $x = - \frac{3}{2} .$ Is this a global minimum of the function? Is there a global maximum? What are the global extrema (if any) if we consider the function restricted to the interval [鈭�3, 3]?

Short Answer

Expert verified

$x = - \frac{3}{2} i s t h e l o c a l m i n i m u m o f t h e g i v e n f u n c t i o n . T h e g i v e n f u n c t i o n d o e s n o t h a v e a n y g l o b a l m a x i m u m . x = - \frac{3}{2} i s t h e g l o b a l m i n i m u m i n t h e i n t e r v a l [- 3, 3] f o r t h e f u n c t i o n g i v e n .$

Step by step solution

Step 1. Given Information.

Given the function: $f (x) = x^{2} + 3 x$ .

Step 2. First-derivative test.

Firstly we need to find the critical points:

$f' (x) = 0 S i n c e, f (x) = x^{2} + 3 x, 2 x + 3 = 0 x = - \frac{3}{2} . I t i s t h e c r i t i c a l p o i n t .$

$N o w, f'' (x) = 2 > 0, t h i s m e a n s x = - \frac{3}{2} i s a l o c a l m i n i m a .$

There is not any global maximum.

If we consider the interval [-3,3] then we get,

$f (- 3) = {(- 3)}^{2} + 3 (- 3) = 9 - 9 = 0 f (3) = 3^{2} + 3 (3) = 9 + 9 = 18 f (- \frac{3}{2}) = {(- \frac{3}{2})}^{2} + 3 (- \frac{3}{2}) = \frac{9}{4} - \frac{9}{2} = - \frac{9}{4} . N o w f r o m a b o v e w e c a n s e e t h a t f (- \frac{3}{2}) i s m i n i m u m . S o, x = - \frac{3}{2} i s t h e g l o b a l m i n i m u m i n t h e i n t e r v a l [- 3, 3] .$