How do you graph the inequality x<0? - Socratic
2016年7月3日 · The region to the left of 0. The type of graph will depend on whether is it a simple number line graph, or on a set of x-y- axes. On a number line graph, we would start with an open circle on 0, because 0 IS NOT included in the solution. Draw a line extending to the left, indicating that x can be any value to the left of 0. On ax-y- grid, we …
How do you solve e^x=0? - Socratic
2016年6月30日 · There is no x such that e^x = 0 The function e^x considered as a function of Real numbers has domain (-oo, oo) and range (0, oo). So it can only take strictly positive values. When we consider e^x as a function of Complex numbers, then we find it has domain CC and range CC "\\" { 0 }. That is 0 is the only value that e^x cannot take. Note that e^(x+yi) = e^x e^(yi) = e^x(cos y+i sin y) We have ...
How do you prove that the function f(x) = | x | is continuous at x=0 ...
2016年3月22日 · That is, the derivative does not exist at x = 0. Calculus . Science Anatomy & Physiology Astronomy ...
When sinx=0, what does x equal? - Socratic
2016年5月24日 · It crosses the x-axis (i.e. it is #0#) at #x = 0, pi,# and #2pi# in the domain #[0,2pi]#, and continues to cross the x-axis at every integer multiple of #pi#. graph{sinx [-10, 10, -5, 5]} And if you click on the graph, you get: So, whenever #sinx = 0#, we have that: #color(blue)(x = pi pm kpi)# for all #k# in the set of integers.
How do you find the limit of #xlnx# as #x->0^-#? - Socratic
2017年6月29日 · There is no limit as x approaches 0 from below since ln x is undefined for negative numbers. Instead, I will demonstrate how to find the right-handed limit, i.e., as x->0^+. Here is a graph: So we should expect the answer to be zero. Now, to do this we can't use the product rule, since the limit of ln x diverges as x->0^+, we have to be more clever. L'HOPITAL'S RULE: You can Google the precise ...
How do you find the limit of #(1-cosx) / x# as x approaches 0?
2016年10月3日 · 0 1-cosx=2sin^2(x/2) so (1-cos x)/x=(x/4) (sin(x/2)/(x/2))^2 then lim_(x->0)(1-cos x)/x equiv lim_(x->0)(x/4) (sin(x/2)/(x/2))^2 = 0 cdot 1 = 0
How do you find the limit of x/sinx as x approaches 0? - Socratic
2016年3月3日 · #lim_{x rarr 0} x/{sin x} = lim_{x rarr 0} 1/{cos x} = 1/{cos 0} = 1/1 = 1#. NOTE The question was posted in " Determining Limits Algebraically " , so the use of L'Hôpital's rule is NOT a suitable method to solve the problem.
How do you know if a solution is "no solution" or "infinite
2014年11月13日 · For an answer to have an infinite solution, the two equations when you solve will equal 0=0. Here is a problem that has an infinite number of solutions. 3x+2y= 12 -6x-4y=24 If you solve this your answer would be 0=0 this means the problem has an infinite number of solutions. For an answer to have no solution both answers would not equal each other. Here is a problem that has no solution. 4x-8y ...
Lim x->0 (tan x- sin x)/x^3=? - Socratic
2017年11月9日 · lim_(x->0) (tanx-sinx)/x^3 = 1/2 Transform the function in this way: (tanx-sinx)/x^3 = 1/x^3(sinx/cosx-sinx) (tanx-sinx)/x^3 = 1/x^3((sinx-sinxcosx)/cosx) (tanx-sinx ...
How can you find the taylor expansion of sin x about x= 0
2015年10月6日 · The Taylor series formula is: sum_(n=0)^N (f^((n))(a))/(n!)(x-a)^n The Taylor series around a = 0 (not x = 0... the question is technically off) is also known as the ...