f(x) = (2x^2 + 5x - 12)/(x + 4)
as given, is not defined when the denominator is 0 ie when x + 4 = 0 or x = -4.
However, 2x^2 + 5x - 12 = (x + 4)*(2x - 3) so that the function can be simplified to (x + 4)*(2x - 3)/(x + 4) = 2x - 3
This has a zero when 2x - 3 = 0 ie 2x = 3 or x = 1.5
∫ f'(x)/[f(x)√(f(x)2 - a2)] dx = (1/a)arcses(f(x)/a) + C C is the constant of integration.
irdk
By factoring I get x-3 divided by x+3
xsquared minus x minus 72
0x-x=0
In such cases, there is usually a discontinuity when the denominator is zero. In other words, solve for:x + 2 = 0
x = 1
This expression factors as x -1 quantity squared.
68.3
2m^2 - 8 -First you should factor out a two. --> 2(m^2-4) -You now have something squared minus something else squared; You have m squared minus 2 squared. Whenever you have something squared minus something squared as you do in this case, there is a simple rule to remember: You can reduce that expression into the quantity of the square root of the first number or variable plus the square root of the second number or variable Times the quantity of the square root of the first number or variable minus the second number or variable squared. --> In the case of your expression: ----> 2(m+2)(m-2)<-----
It is the straight line through the points (0, -1) and (1, 0).
∫ f'(x)/[f(x)√(f(x)2 - a2)] dx = (1/a)arcses(f(x)/a) + C C is the constant of integration.
3x squared minus 25x minus 28
∫ f'(x)/√(a2 - f(x)2) dx = arcsin(f(x)/a) + C C is the constant of integration.
2x squared minus 4
-2, 1.74 and 0.46
∫ f'(x)/( q2f(x)2 - p2) dx = [1/(2pq)ln[(qf(x) - p)/(qf(x) + p)]