What is [math] X^2 [/math] times X?
I assume you’re writing with a tablet that doesn’t show exponents or use superscript… Mine doesn’t allow them…In order to show exponents try this on your tablet or phone for greater understanding: X^2, will allow the viewer to determine you mean X squared.When multiplying numbers or variables with the same base, keep the base and add the powers together: X^2=X times X. So the problem breaks down to, what is X times X times X. This is written as X^3, or X cubed. We know this is true because 4 x 4 x 4 equals 64, and 4^3 (4 cubed) equals 64…Hope this helps…
How do I solve the inequality [math]x (x-1) (x+2) > 0[/math]?
The question is x(x-1)(x+2)>0Firt ignore the '>' sign:so: x(x-1)(x+2)=0Since 3 numbers are multiplied to give 0, then each may potentially be zero, so: x = 0 x - 1 = 0, which means x = 1 x + 2 = 0, which means x = -2So, you have 3 numbers -2, 0, 1 Now, draw a straight line and mark these 3 points on the line. You can see now that it these numbers have divided the line into 4 different regions. Region 1 is (-inf, -2) -- that is all numbers to the left of -2.Region 2 is (-2,0) --- that is all numbers between -2 and 0Region 3 is (0,1) --- that is all numbers between 0,1Region 4 is (1,inf)--- that is all numbers to the right of 1 Your strategy now, is to test a number inside of each of these 4 regions. Basically you pick a number inside, but not the end values. You substitute the number into the original inequality for x, and see which ones are true.Hence:Region 1: (-inf, -2), I can choose -5 since its in the region to the left of -2.I get: (-5)(-5-1)(-5+2) >? 0 I used '?' because I am not sure if it is trueso: (-5)(-6)(-3) >? 0 Finally -90 >? 0 is false. SO Mark this region with an X on your lineRegion 2: (-2,0), I can choose -1 since its between -2 and 0.I get: (-1)(-1-1)(-1+2) >? 0 I used '?' because I am not sure if it is trueso: (-1)(-2)(1) >? 0 Finally 2 >? 0 is True. SO Mark this region with an check mark on your lineRegion 3: (0,1), I can choose 0.5 since its between 0 and 1.I get: (0.5)(0.5-1)(0.5+2) >? 0 I used '?' because I am not sure if it is trueso: (0.5)(-0.5)(2.5) >? 0 Finally -0.625>? 0 is false. SO Mark this region with an X on your lineFinally: Region 4: (1,inf), I can choose 5 since its in the region to the right of 1.I get: (5)(5-1)(5+2) >? 0 I used '?' because I am not sure if it is trueso: (5)(4)(7) >? 0 Finally 140 >? 0 is true. SO Mark this region with a check on your lineFinally: We have 2 regions that were true (check marks)Regions 2 and 4: (-2,0) and (1,inf)We write them together using the U (for union)Final answer: (-2,0) U (1,inf)
When I want to multiply x by x, is the result [math]2x[/math] or [math]x^2[/math]?
Firstly, this is Algebra. The product of two like terms, which in this case is x multiplied by x, is x squared. They are being multiplied and are the same term, so of course will be squared.Secondly, 2x is two lots of x. It means x + x. Think of it as 1+1.
For what value of x does the function f(x) = (x - 2)(x - 3)^2 have a relative maximum?
take the derivative of the function f'(x) = (x-3)² + 2(x-3)(x-2) f'(x) = (x-3)[x-3+2x-4] f'(x) = (x-3)(3x-7) set it equal to 0 f'(x) = (x-3)(3x-7)=0 x= 3, 7/3 take the second derivative f''(x) = (3x-7) + 3(x-3) f''(x) = 3x-7+3x-9 f''(x) = 6x-16 plug in the values of the critical point from the first derivative in the second derivative f''(3) = 6(3) - 16= 2>0, since it greater than 0 it concave up and is a minimum f''(7/3) = 6(7/3) -16= -2<0, less then 0, so concave down is a maximum
What is the domain of ln (x - x^2)?
The domain of the function is the set of values that it can accept as x in this case. Noting that the logarithm function is only defined for positive real numbers (that is, all real numbers strictly non-zero),ln(z) is defined for all z in R with z > 0.Using this with z = x - x^2,x - x^2 > 0 → x^2 - x < 0→ x(x-1) < 0→ 0 < x < 1So as long as x is in the interval (0,1), your function is defined.I know that Alex already answered this, so credit to his answer. Just providing some more detailed reason.Kind regards,Zane Heyl