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# Calculus Problems?

Topic: How to write a function rule for a graph
June 27, 2019 / By Candis
Question: I have some questions about some calculus problems. If anyone could explain to me how they are done it would be much appreciated: A) If f(x) = x^3, find an expression for d/dx [f(g(x))]. B) Find dy/dx for sin(xy) = y. C) Consider the curve given by y^3 + 5x^2y - 18 = 0. Write an equation for the line tangent to the curve at the point (1,2). D) If x and y are both differentiable functions of t, and xy = 20, find x'(t) when y'(t) = 10 and x = 2. E) Given that t, k, and a are constants, and that f(x) = a - 2kx, find f'(t). F) If f'(a) does NOT exist, which of the following MUST be true? If false, show an example of a graph where it is false. 1. f(x)is discontinuousat x = a 2. the limit as x-->a of f(x) does not exist 3. f(x) has a vertical tangent at x = a 4. f(x) has a "hole" at x = a 5. none of the above is necessarily true. Thanks for any help you can give me!!!

Ambrosia | 10 days ago
A: This is a simple application of the chain rule: d(f(g(x)))/dx = f'(g(x)) g'(x) = 3g(x)²g'(x) B: Here we differentiate implicitly, and then solve for dy/dx: sin (xy) = y d(sin (xy))/dx = dy/dx cos (xy) d(xy)/dx = dy/dx cos (xy) (y + x dy/dx) = dy/dx y cos (xy) + x cos (xy) dy/dx = dy/dx y cos (xy) = dy/dx - x cos (xy) dy/dx y cos (xy) = (1 - x cos (xy)) dy/dx dy/dx = y cos (xy)/(1 - x cos (xy)) C: First, we use implicit differentiation to find the slope, then we write the line in point-slope form: 3y² dy/dx + 10xy + 5x² dy/dx = 0 (3y²+5x²) dy/dx = -10xy dy/dx = -10xy/(3y²+5x²) Substituting (1, 2): dy/dx = -20/(12+5) = -20/17 The tangent line is therefore: y-2 = -20/17 (x-1) y-2 = -20x/17 + 20/17 y = -20x/17 + 54/17 D: Again, using implicit differentiation: xy=20 d(xy)/dt = d(20)/dt y dx/dt + x dy/dt = 0 dx/dt = x/y dy/dt From our original equation: xy=20 y=20/x So: dx/dt = x²/20 dy/dt Substituting: dx/dt = 4/20 * 10 = 2 E: f'(t) = -2k. This is about the simplest derivative in the world F: None of the above is necessarily true. In fact, a single counterexample refutes all four: the function f(x) = |x| This function: Has a limit at 0 ([x→0+]lim f(x) = [x→0+]lim f(x) = f(0) = 0) Is continuous at 0 (per the above) Does not have a hole at 0 (since f(0) is defined -- specifically, f(0)=0). Has NO tangent at 0 And is not differentiable at 0 (since [h→0]lim (f(0+h) - f(0))/h = [h→0]lim |h|/h, but [h→0+]lim |h|/h = 1 and [h→0-]lim |h|/h = -1, and 1≠-1).
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We found more questions related to the topic: How to write a function rule for a graph

Originally Answered: Two Pre-Calculus/Calculus problems I can't figure out, can you help?
A = - C B = D Subsitute 4 (-C) + 3D + 2C = 7 (-C) + 4(D) + 6C = -6 - 2 C + 3 D = 7 5 C + 4 D = -6 8 C -12 D = -28 15 C + 12 D = - 18 23 C = - 46 C = -2 -2 (-2) + 3D = 7 3 D = 3 D = 1 A = 2 B = 1 2.) Complete the square 2 (x^2 + 7/2 x + 11/2) Add (b/2)^2 to complete the square Balance the equation by subtracting 49/16 2 (x^2 + 7/2 x + 49/16 + 11/2 - 49/16) 2 [(x + 7/4)^2 + 39/16]
Originally Answered: Two Pre-Calculus/Calculus problems I can't figure out, can you help?
1. You can solve the first two equations for single variables and substitute into the second two and see where that gets you, but a faster method is rewriting the system as a matrix and reducing it to reduced row-echelon form. | 1, 0, 1, 0 | 0 | | 0, 1, 0, -1 | 0 | | 4, 0, 2, 3 | 7 | | 1, 4, 6, 0 | -6 | Saving you the details, this matrix reduces to: | 1, 0, 0, 0 | 2 | | 0, 1, 0, 0 | 1 | | 0, 0, 1, 0 | -2 | | 0, 0, 0, 1 | 1 | The Linear Algebra Toolkit can show you the steps: http://www.math.odu.edu/~bogacki/lat/ 2. First, simplify the equation: p[(x+q)^2 + r] = p(x^2 + 2xq + q^2 + r) = px^2 + 2pqx + pq^2 + pr And compare coefficients to the given equation. We can see that p=2, 2pq = 7, and pq^2+pr=11. Substitute the new value of p into the equations: 4q = 7, 2q^2 + 2r = 11 Simplify: q = 7/4, q^2 + r = 11/2 Substitute the new value of q into the last equation and simplify: (7/4)^2 + r = 11/2 49/16 + r = 11/2 r = 39/16 Substitute these values into the first equation: 2[(x + 7/4)^2 + 39/16]
Originally Answered: Two Pre-Calculus/Calculus problems I can't figure out, can you help?
I'll get you started, 1) Use equations 1 and two to solve the variables in terms of each other. For example, from equation 2, B=D. 2) This involves completing the square, 2x^2+7x+__ can be factored into some p[(x+q)^2 + r]. First thing you should do is factor out that 2 to make the coefficient of x^2 be 1. What ever you need to add to that 11 to complete the square is r. Hope that leads you in the right direction :)

Walt
A) Use the chain rule. Recall that the chain rule states: [f(g(x))]' = f'(g(x)*g'(x) B) Implicit differentiation. See C for example. C) Here you should take an implicit derivative to find dy/dx. In this case, differentiating implicity with respect to x gives: 3y^2 dy/dx + 10xy + 5x^2 dy/dx = 0 Solve this for dy/dx, then plug in (1,2) to get the slope of the tangent line. D) Take the derivative of both sides of the equation. Plug in appropriate values. E) Take the derivative, evaluate at t. F) Consider the absolute value function at a = 0.
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Originally Answered: Solve the following pre-calculus problems and explain?
Here is help with 2.2: To solve for the vertex when written in the form f(x) = ax^2 + bx + c, we want to change the equation to the form f(x) = a(x-h)^2 + k because when written in this form (h,k) is the point of the vertex. Using the equation x^2 - 6x + 8. In order to change the form of this equation we want to use a method called "complete the square." x^2 - 6x +Y + 8 -Y. Since we are both adding Y and subtracting Y we have not changed this equation in any way. Now we want to pick the value of Y so that the beginning of the equation x^2 - 6x + Y is a perfect square polynomial [will be able to reduce to (a + b)^2]. In this case we want Y to be 9. x^2 - 6x + 9 reduces to (x-3)^2. So that means our original equation x^2 - 6X + 9 + 8 - 9 is now (x-3)^2 + 8 - 9 or (x-3)^2 - 1. The equation is now in the form a(x-h)^2 + k so we now know that the vertex is located at (3,-1).
Originally Answered: Solve the following pre-calculus problems and explain?
to clean up this, write it in the variety x^4=-80 one=80 one(cos(pi)+i sin(pi))=81cis(pi), then use De Moivre's Theorem: if z=r cis(a), then z^n=r^n cis(na)). hence, x^4=z=80 one cis(pi), so one answer is x=z^(one million/4)=80 one^(one million/4) cis(pi/4)=3 cis(pi/4)=sqrt(3)+i sqrt(3). it is only certainly one of four innovations, in spite of the shown fact that. the different 3 innovations are calmly spaced around a circle of radius 3 concentrated on the beginning place, so the finished set of innovations in oblong variety is {sqrt(3)+i sqrt(3), sqrt(3)-i sqrt(3), -sqrt(3)+i sqrt(3), -sqrt(3)-i sqrt(3)}.

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