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Misha Larkins [42]
3 years ago
9

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

Mathematics
1 answer:
shutvik [7]3 years ago
8 0

Answer:

<em>given function has 2 minimums</em> - \frac{9}{4}  and  \frac{9}{4}  

Step-by-step explanation:

<u><em>Step 1.</em></u> g'(x) = 4x³ - 10x

<u><em>Step 2.</em></u> 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} }

<u><em>Step 3.</em></u> 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} }

<u><em>Step 4.</em></u>

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

If x = - \sqrt{\frac{5}{2} } , g(x) has <em>minimum</em> 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 <em>minimum</em> 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

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