For which pair of functions f(x) and g(x) below, will the lim x->∞ (f(x))/(g(x))=0 ?
Well, you write : "I really don't understand this question" ... there is nothing to understand, so to say : it is asked to form : f(x)/g(x) in each case and to answer if Y/N : lim x->∞ (f(x))/(g(x)) = 0 the kernel of the question is : when : x ---> +oo if g(x) is more powerful, growing towards +oo faster than f(x) then : lim x->∞ (f(x))/(g(x)) = 0 if f(x) is more powerful then : lim x->∞ (f(x))/(g(x)) = +oo so here : you must use the rules : 1- a exponantial function is more powerful than any polynomial or monomial x^n for any n 2- a polynomial function is more powerful than a log function therefore : (A) f(x) = e^x, g(x)= x^2 ==> lim x->∞ e^x /x^2 = +oo (rule 1- ) ==> NO (B) f(x) = e^x, g(x)= ln x ==> lim x->∞ (e^x /x) * (x / lnx) = +oo * +oo = +oo ==> NO (C) f(x)= ln x, g(x)= e^x ==> lim x->∞ (lnx /x) * x/ e^x = 0 * 0 = 0 ==> YES (E) f(x)= 3^x g(x)= 2^x ==> lim x->∞ f(x)/g(x) = lim x->∞ (3/2)^x = +oo ==> NO unique YES is (C) <--- ANSWER hope it' ll help !!
Use the graph below to estimate the indicated derivatives, or state that they do not exist.?
Since this IS homework (or a quiz) - that is what "webwork" is for - I will not give a straight answer, rather I will attempt to explain what you are supposed to know. ----------------- First, you have studied the chain rule: (dy/dx) = (dy/du)×(du/dx) or (d/dx)(F(G(x))=F'(G(x))×G'(x) ----------------- All they are asking you to do is plug in x=1 and x=2 and x=3. Looking at the graph (it is a very clear graph, but you might want to mark it off in 1/4 units instead of whole units): What is G(1)=? G(2)=? G(3)=? ---------- Second, you also need to know that the derivative is the function which gives the slope at a point. G'(1) is the slope of the graph y=G(x) AT THE POINT x=1. ---------- Since these graphs are pieces of straight lines that part is junior high school work: G'(1)=? G'(2)=? G'(3)=? F'(G(1))=? F'(G(2))=? F'(G(3))=?. So multiply to get h'(1), and h'(2) and h'(3). Note: You are NOT asked for F'(2), which is the only (non-end-)point were F' is undefined. h'(x) is defined for all x between 0 and 4, so the answer is not "dne". for this problem. P.S. There is no REAL problem getting the image. Any problem is a result of an "advanced" browser (with no real intelligence) which thinks it knows better than the user. (Which may be unfortunately true in too many cases.) - Can you tell that I don't user Microsoft products?
How do I find the points where f(x) is not differentiable?
Case 1 :Firstly check the continuity of that function. The function is not differentiable at any point if the function is discontinuous at that point..Because the curve doesn't exist at the point of discontinuity, so not differentiable at that point.Case 2 :In some specific cases, a function may not be differentiable at the point where the function is continuous. Such points are breaking point.** Breaking points are such points where there is not possible to draw a tangent to the curve at that point .For an example y = |x| is a function which is continuos at (0,0) but not differentiable. For the function y = |x| , (0,0) is a breaking point.** And if you find the left hand limit, L f ' (P) and right hand limit, R f ' (P) of a function f(x) at point P, The function will be differentiable if (i) L f ' (P) and R f ' (P) are defined.(ii) L f ' (P) = R f ' (P) ..Thus, You can also check differentiability of a function using the concept of limit.
How do I determine the equation of the tangent to the curve [math] y=2x-x^2 [/math] that passes through point [math] (2,9) [/math]?
There are actually two such tangents:Let y = mx + n be the tangent we're looking for, i.e. we have to determine m and n.By definition of tangent, the line y = mx + n is a tangent to the given curve if they interesect in exactly one point. So let's equate the two equations to calculate intersection(s):[math]y = mx + n = 2x - x^2,[/math][math]x^2 + (m-2)x + n = 0,[/math][math]x = \frac{m-2}{2}\pm\sqrt{\left(\frac{m-2}{2}\right)^2-n}.[/math]If the radicand (the "thing" under the square root) is positive, there will be two intersections. If it is negative, there will be no intersection. If it is zero, there will be one intersection, and the line will indeed be a tangent.So we get:[math]\left(\frac{m-2}{2}\right)^2-n=0,[/math][math]n=\left(\frac{m-2}{2}\right)^2.[/math]Now we can use the fact that the tangent passes through (2,9):[math]9=2m+n=2m+\left(\frac{m-2}{2}\right)^2=2m+\frac{\left(m-2\right)^2}{4},[/math][math]36=8m+4-4m+m^2,[/math][math]m^2+4m-32=0,[/math][math]m=-2\pm\sqrt{4+32}=-2\pm\sqrt{36}=-2\pm 6,[/math][math]m=4\ \text{or}\ m=-8.[/math]For m = 4, we get[math]n=\frac{(2-4)^2}{4}=\frac{4}{4}=1,[/math]and for m = -8, we get[math]n=\frac{(2-(-8))^2}{4}=\frac{100}{4}=25.[/math]The two solutions are therefore y = 4x + 1 and y = -8x + 25.
A polynomial leaves remainder [math]2[/math] when divided by [math]x-1[/math] and remainder [math]1[/math] when divided by [math]x-2[/math]. If the polynomial is divided by [math](x-1)(x-2)[/math], then what would be the remainder?
By remainder theorem, “When f(x) is divided by (x−1), the remainder is 2” can be translated to:-[math]f(1)=2[/math] ….(Equation-1)Similarly, we have:-[math]f(2)=1[/math].......(Equation-2)When [math]f(x)[/math] is divided by [math](x−1)(x−2)[/math], then basically we have:-[math]f(x)=(x−1)(x−2)[/math]Quotient + Remainder……(Equation-3)Since the degree of the remainder must be one lower than that of the divisor, [math]=2[/math] from [math](x−1)(x−2)[/math]], the remainder can have degree = 1 (or lower) only. The remainder should then take the form [math]ax+b[/math], which is a general expression of degree 1 (or lower if a = 0) in x. Therefore, (3) becomes:-[math]f(x)=(x−1)(x−2)Q(x)+(ax+b)[/math].....(Equation-4)(1) and (2) can be used to find the values of [math]a[/math] and [math]b[/math] from (4).f(1) = a+b and f(2) = 2a+b…. (From equation 1,2 and 4)a+b = 2 and 2a+b=1Solving the given equation,a= -1 and b=3Thus, Remainder = ax+b = -x+3
Calculus: If h(x) = f(g(x)), then h'(1) = ...?
The instructions say let f and g be differentiable functions such that f(1) = 4 g(1)=3 f '(3)=-5 f '(1)=-4 g '(1)=-3 g '(3)=2 I don't know why they gave the f ' and g ' of 3, unless to simply confuse. I thought I knew how to do this type of problem, but I keep getting 36 as an answers, when the only possible ones are -9, 15, 0, -5, -12
Let f(x) = 3x - 4 and g(x) = −x2?
f(x) = 3x - 4 and g(x) = −x^2 (f + g)(x) = f(x) + g(x) = (3x - 4) + (-x^2) = - x^2 + 3x - 4 (f - g)(x) = f(x) - g(x) = (3x - 4) - (-x^2) = x^2 + 3x - 4 (f * g)(x) = f(x) * g(x) = (3x - 4) * (-x^2) = -3x^3 + 4x^2 (f / g)(x) = f(x) / g(x) = (3x - 4) / (-x^2) = 4/x^2 - 3/x (f o g)(x) = f( g(x) ) = 3 ( g(x) ) - 4 = 3 ( -x^2 ) - 4 = -3x^2 - 4 (f • g)(0) ? if (f * g)(0) = -3(0)^3 + 4(0)^2 = 0 if (f o g)(0) = -3(0)^2 - 4 = - 4