Question

Pls help asap whats the local min value of the function below?

Answer 1

Answer:

given function has 2 minimums - $\frac{9}{4}$  and  $\frac{9}{4}$  

Step-by-step explanation:

Step 1. g'(x) = 4x³ - 10x

Step 2. Find find the critical points:

4x³ - 10x = 2x(2x² - 5) = 0

$x_{1}$ = - $\sqrt{\frac{5}{2} }$ , $x_{2}$ = 0 , $x_{3}$ = $\sqrt{\frac{5}{2} }$

Step 3. g'(x) > 0 :  - $\sqrt{\frac{5}{2} }$ < x < 0  or  x > $\sqrt{\frac{5}{2} }$

g'(x) < 0 :   x < - $\sqrt{\frac{5}{2} }$   or   0 < x < $\sqrt{\frac{5}{2} }$

Step 4.

If x ∈ ( - ∞ , - $\sqrt{\frac{5}{2} }$ ) , g(x) is decreasing ;

If x = - $\sqrt{\frac{5}{2} }$ , g(x) has minimum value ;

If x ∈ ( - $\sqrt{\frac{5}{2} }$ , 0 ) , g(x) is increasing ;

If x = 0 , g(x) has maximum value ;

If x ∈ ( 0 , $\sqrt{\frac{5}{2} }$ ) , g(x) is decreasing ;

If x = $\sqrt{\frac{5}{2} }$ , g(x) has minimum value ;

If x ∈ ( $\sqrt{\frac{5}{2} }$ , ∞ ) , g(x) is increasing .

⇒ at ( - $\sqrt{\frac{5}{2} }$ , - $\frac{9}{4}$ ) and at ( $\sqrt{\frac{5}{2} }$ , $\frac{9}{4}$ ) , g(x) reaches its minimum