Find interval of increase and decreaseSo im supposed to find the interval of decrease and increase here. Ive gotten up to taking the derivative which is − 4 x ( x 2 − 1 ) and then setting it to 0 i got (-1,0,1) Im lost at what to do now?Im supposed to take it for this below: f ( x ) = 7 + 2 x 2 − x 4

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Find interval of increase and decreaseSo im supposed to find the interval of decrease and increase here. Ive gotten up to taking the derivative which is   −  4  x  (      x    2    −  1  ) and then setting it to 0 i got (-1,0,1) Im lost at what to do now?Im supposed to take it for this below:  f  (  x  )  =  7  +  2      x    2    −      x    4

Paulina Horne · Accepted Answer

Step 1The derivative is continuous everywhere; that means that it cannot change signs without going through 0, by the Intermediate Value Theorem. Since the only places where the derivative is zero are at   x  =  −  1,   x  =  0, and   x  =  1, that means that those are the only places where the derivative could, possibly, change signs.Why do we care about where the derivative could change signs? Because the sign of the derivative tells you whether the original function is increasing or decreasing.So... what is the sign on the derivative on   (  −  ∞  ,  −  1  )? Plugging in   x  =  −  2 you get a positive number. Since it cannot change signs on this interval, the derivative must be positive on all of   (  −  ∞  ,  −  1  ). That tells you that the original function, f(x), is increasing on the interval   (  −  ∞  ,  −  1  ] (you get to include -1 as well).Step 2To see what happens next, you should determine whether f′(x) is positive or negative on (-1,0); on (0,1), and on   (  1  ,  ∞  ). This can be done by either using a &quot;test point&quot;, or analyzing the derivative. For example, considering (-1,0). Since       f    ′    (  x  )  =  −  4  x  (      x    2    −  1  )  =  −  4  x  (  x  −  1  )  (  x  +  1  ), if we plug in a negative number for x then -4x will be positive. Because x is between -1 and 0,   x  +  1 will be positive. And since x is already negative,   x  −  1 will be negative as well. So       f    ′    (  x  )  =  −  4  x  (  x  −  1  )  (  x  +  1  ) will be a product of two positive numbers and a negative number, so f′(x)  is negative on (-1,0). That means that f(x) is decreasing on [-1,0].Now do the same for the remaining intervals of constant sign for the derivative.

Gorlandint · Accepted Answer

Explanation:You have stationary points with you(where derivative   =  0).Now, check for each interval created by these points whether f′(x) is positive or negative in that interval.In your problem,       f    ′    (  x  )  =  −  4  x  (  x  −  1  )  (  x  +  1  ) is positive for   x  &amp;lt;  −  1, negative for   −  1  &amp;lt;  x  &amp;lt;  0, positive for   0  &amp;lt;  x  &amp;lt;  1 and negative for   x  &amp;gt;  1. The interval where derivative f′(x) is positive, f(x) is increasing and decreasing where f′(x) is negative.

Im supposed to find the interval of decrease and increase here. Ive gotten up to taking the derivative which is -4x(x^2-1) and then setting it to 0 i got (-1, 0, 1). What to do now? Im supposed to take it for this below: f(x)=7+2x^2-x^4

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