MHB Troubleshooting Inequality: Can't Seem to Solve the Last One

AI Thread Summary
The discussion focuses on solving an inequality involving a function f(x) and determining the correct intervals where f(x) is less than zero. The user is struggling to identify the last interval needed for the solution. It is clarified that the values -2 and 3 cannot be included in the solution since f(x) must be strictly less than zero. The correct intervals are identified as (-∞, -2) and (3, ∞). The conversation emphasizes the importance of understanding strict inequalities in interval notation.
ahbm
Messages
1
Reaction score
0
Trying to get the last one, but not able to do it. What am I missing. I know that I can't but [-2,infinity] or [3,infinity] because for -2, and 3 y=0 and is not > 0. Any help appreciated.
 

Attachments

  • Calculus.png
    Calculus.png
    14 KB · Views: 114
Mathematics news on Phys.org
Is g(-2) > 0? No, so that part of the interval will be [math]( -\infty, -2)[/math].

-Dan
 
Also, because the question requires that f(x) be strictly less than 0, f(x)< 0 you cannot include x= -2 or x= 3. f(x)< 0 is true for (-\infty, -2)\cup (3, \infty).
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Back
Top