Directions: Please write your solution on the space provided and boxed your final answer. *Introduction to Derivatives (x²+2x-3 ,if x #1 if x 1 Using the 3 conditions, show that f(x) = x-1 is both continuous 4 and differentiable at x = 1.
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- Let f(x,y)= 6-3xy+x2 a) The function f has only one stationary point. Find it. b) Is the stationary point of f a local maximum, or local minimum, or a saddle point? Justify your answer.Find all relative extrema of the function. (If an answer does not exist, enter DNE.) f(x) = 5x - 60x^1/3 Relative Maximum (x,y) =? Relative Minimum (x,y) =?Q6.) Does ƒ(x) = x3 + 2x + tan x have any local maximum or minimum values? Give reasons for your answer.
- First-class postage is $0.44 for the first ounce (or any fraction thereof) and $0.17 for each additional ounce (or fraction thereof) up to a maximum weight of 3.5 ounces. (A) Write a piecewise definition of the first-class postage P(x) for a letter weighing x ounces. (B) Graph P(x) for 0<x≤3.5. (C) Is P(x) continuous at x=2.5? At x=3? Explain.lim(x,y) aproaches (0,0) (3x3y)/(x6+y2)Help, need to double check answer. Please answer completely. 1. Using the Second Derivative Test, find and classify all the relative extrema and saddle points of this function f(x,y) = 3xy(x+2y-6)
- Find all relative extrema of the function. (If an answer does not exist, enter DNE.)f(x) = x4 − 32x + 8relative maximum (x, y) = relative minimum (x, y) =Continuity At what points of ℝ2 are the following functions continuous? ƒ(x, y) = x2 + 2xy - y3f the first derivative changes from positive to negative at point c, what does this mean? c is a global maximum c is a local maximum c is a local minimum c is a global minimum
- Consider the function ƒ(x, y) = x2 + y2 + 2xy - x - y + 1 over the square 0 <=x <= 1 and 0<= y <=1. a. Show that ƒ has an absolute minimum along the line segment 2x + 2y = 1 in this square. What is the absolute minimum value? b. Find the absolute maximum value of ƒ over the square.Fast pls solve this question correctly in 5 min I will give you like for sure. Given f(x,y) = 5x3-3/2y2-15xy, find and label all points in the xy-plane where a saddle point or relative extrema exists.Find the values of h, k, and a that make the circle (x - h)2 + (y - k)2 = a2 tangent to the parabola y = x2 + 1 at the point (1, 2) and that also make the second derivatives d2y/dx2 have the same value on both curves there. Circles like this one that are tangent to a curve and have the same second derivative as the curve at the point of tangency are called osculating circles.