Calculus: Derivatives Problem #2

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The discussion revolves around understanding the behavior of the function y=2x^3+24x-18 in relation to its derivative, y'=6x^2+24. The key point is that the function is increasing for all values of x because the derivative is always positive, despite the function's output being negative for certain x values. The confusion arose from equating the function's value being negative with the derivative's positivity. Ultimately, it was clarified that a function can increase even when its values are negative. This highlights the importance of differentiating between the function's output and its rate of change.
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[SOLVED] Calculus: Derivatives Problem #2

Homework Statement


The graph of y=2x^3+24x-18 is


Homework Equations


y '=6x^2+24
y ''=12x

The Attempt at a Solution


The answer to this is "increasing for all values of x"
But I want to know why...if substituting -1 for x will make the problem a negative.
 
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What's the question? If you want to find where y is increasing it's where is y' positive. You don't think y' is positive for all x?
 
because when you plug in -1 for y=2x^3+24x-18 the end result is a negative not positive.

The other choices to the questions are:
b.decreasing for all values of x
c.only increasing for values of x on the interval (-infinity,-2)U(2,+infinity)
d.only increasing for values of x on the interval (-2,2)
e. only decrasing for values of x on the interval (-infinity,-2)
 
NVM, TOTAL IDIOT HERE... I know what you're saying now. Thanks.
 
The question isn't where y is negative, it's where y in increasing. y can be increasing even if it's negative.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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