<span>3down votefavorite1Find minimum and maximum value of function <span>f(x,y)=3x+4y+|x−y|</span> on circle<span>{(x,y):<span>x2</span>+<span>y2</span>=1}</span>I used polar coordinate system. So I have <span>x=cost</span> and <span>y=sint</span> where <span>t∈[0,2π)</span>.Then i exploited definition of absolute function and i got:<span>h(t)=<span>{<span><span>4cost+3sintt∈[0,<span>π4</span>]∪[<span>54</span>π,2π)</span><span>2cost+5sintt∈(<span>π4</span>,<span>54</span>π)</span></span></span></span>Hence i received following critical points (earlier i computed first derivative):<span>cost=±<span>45</span>∨cost=±<span>2<span>√29</span></span></span>Then i computed second derivative and after all i received that in <span>(<span>2<span>√29</span></span>,<span>5<span>√29</span></span>)</span> is maximum equal <span>√29</span> and in <span>(−<span>45</span>,−<span>35</span>)</span> is minimum equal <span>−<span>235</span></span><span>
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Please note that your x^3/4 is ambiguous. Did you mean (x^3) divided by 4
or did you mean x to the power (3/4)? I will assume you meant the first, not the second. Please use the "^" symbol to denote exponentiation.
If we have a function f(x) and its derivative f'(x), and a particular x value (c) at which to begin, then the linearization of the function f(x) is
f(x) approx. equal to [f '(c)]x + f(c)].
Here a = c = 81.
Thus, the linearization of the given function at a = c = 81 is
f(x) (approx. equal to) 3(81^2)/4 + [81^3]/4
Note that f '(c) is the slope of the line and is equal to (3/4)(81^2), and f(c) is the function value at x=c, or (81^3)/4.
What is the linearization of f(x) = (x^3)/4, if c = a = 81?
It will be f(x) (approx. equal to)
Answer:
umm, 3/4
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bru
Step-by-step explanation:
Answer:.
Step-by-step explanation:.
