Absolute minimum? HELP!! f(x)=2x^3 + 6x^2 -144x +11 , [-6,5]?

Question

i tried everything!!

Answer 1

There is an interval given, -6
If you're in calculus, take the derivative, and find critical points by setting it equal to 0.
f'(x) = 0= 6x^2 + 12x -144. ...I suggest pulling a 6 out.
0 = x^2 + 2x - 24 = (x + 6)(x-4), x = -6 and x = 4. Plug these X-values into the original equation.
If X = -6, Y = 659. If X = 4, Y = -341. Since X = 4 gives you the lower answer, the absolute minimum is at the point (4, -341).


...But this only works if you know differential calculus. Otherwise, I would recommend plugging in all the points from -6 to 5 into the function, drawing out a graph, and getting an answer there. Or, save time and use a graphing calculator.

Answer 2

I would say(4, -341) to prove it u have to use derivatives where the derivative is zero between -6 and 5 is where the function has a minimum

Answer 3

For a cubic function there is no minimum value since its arms stretches towards positive and negative infinity. However, you can find the minimum value of one of the curves as the graph curves down. Using a graphing calculator, you would get, (4, -341)

Answer 4

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Answer 5

I don't know I just want points.