Absolute Minimum

Calculus Level 2

What is the absolute minimum value of the function $2x^{3}-3x^{2}-12x+1$ in the interval $[0,4]$ ?

-2 6 1 -1 -8 -19 -6 0

1 solution

Divyansh Chaturvedi
May 18, 2015

Just differentiate the function and equal it to 0. We get , 6x^2 - 6x - 12 = 0 Hence, x^2 - x - 2 = 0 (x-2)(x+1) = 0 since 0<= x <= 4 So x = 2.

Putting it in the function 16-12-24+1 = -19

You also need to check the value of the function at the endpoints.

Since it could be the case that at $x=0$ or $x=4$ it is less than the local min.

In this case it's not. But it's important to check.

Isaac Buckley - 5 years, 11 months ago

you have to considerate Buckley's comment and check that the function has a local minimum in x=2,you haven't finished of checking it. For example, its second derivative in x= 2 is greater than 0 or seeing that the first derivate is less than 0 to the left of x=2(the function is decreasing to the left of x=2) and greater than 0 to the right of x=2( the function is increasing to the right of x=2)

Guillermo Templado - 5 years, 8 months ago

Absolute Minimum

1 solution

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