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18. [-f'l Points] DETAILS OSCALC1 3.6.259. The depth (in feet) of water at a dock changes with the rise and fall of tides. The depth is modeled by the Function D(t), where t is the number of hours after midnight. om = 2 sin(r 7 7\") + 7 5 5 Find the rate [in ftfhr) at which the depth is changing at 6 a.m. (Round your answer to one decimal place.) Em \f1. [1/2 Points] DETAILS PREVIOUS ANSWERS OSCALC1 3.7.262. Consider the graph of y = f(x). -4 -3 -2 -1 1 2 3 -1 -2 -3 (a) Sketch the graph of y = f-1(x). N 2 N 1 1 -3 -2 -1 2 3 -3 -2 -1 1 2 3 4 -4 -3 -2 -1 2 3 -4 -3 -2 -1 1 2 3 -1 -1 -1 1 -2 -2 -2 -3 -3 -3 LO -4 O -4 (b) Use part (a) to estimate (f-1)'(1).Step 3 of 4 (c) Lastly, we use f (y) = v 100 - y to find df at y = A(3). dy The inverse function theorem tells us that for an invertible, differentiable function x = f(y), df - 1 1 wherever f'(f (v) ) = 0. dy F' ( f - 1 ( ) ) ) Since y = f(3) = 100 - 32 = 91 , we have the following. df - 1 1 dy F' ( f - () ) ) df - 1 dy ly = (3) = 100 - H = Submit Skip (you cannot come back)4. [0/1 Points] DETAILS PREVIOUS ANSWERS OSCALC1 3.7.273. Find (F- 1)'(a). f( x) = tan-1(x) + 3x2, a = 0 (F-1)' (a) = 0 X 5. [1/2 Points] DETAILS PREVIOUS ANSWERS OSCALC1 3.7.274. Consider the function. 10 f ( x ) = 1 + x2 (a) Find the slope of the tangent line to its inverse function fat point P(5, 1). (b) Find the equation of the tangent line to the graph of fat point P(5, 1). (Let x be the independent variable and y be the dependent variable.) X + 2 X7. [1/5 Points] DETAILS PREVIOUS ANSWERS OSCALC1 3.7.283.TUT.SA. This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise Find Jy y = (1 + cot-1(x)) 4 Step 1 of 4 To determine - for the given function, first notice that y = (1 + cot"(x))* can be viewed as a composition of two functions with an "outside" function, f(x), and an "inside" function, g(x), where y = Alg(x)) = (1 + cot->(x))4. Therefore, we will start with the chain rule to determine the derivative. Recall that the chain rule tells us the following for the composition y = f(9(x)). ay = f'(g(x) ) g'(x) dx In considering the outside function, which are possible choices for f(x)? (Select all that apply.) f ( x ) = cot-1 (x) V D A( X ) = X 4 V D A( x ) = ( 1 + x ) 4 f ( x ) = 1+ cot -1(x) f ( x ) = 1 + x Step 2 of 4 We choose f(x) = x# as our outside function since finding f'(x) will not require another application of the chain rule. This leaves only one choice for the inside function g(x). 4(1 + cot- 1 (x) ) 3 g ( x ) = 1 + x 2 X cot- (x) + 1 Step 3 of 4 We have f(x) = x4 and g(x) = 1 + cot-1(x), so we next determine the derivative of each function. F' ( x ) = and g'( x ) = = (1 + cot-1(x)) = Submit | |Skip (you cannot come back)9. [0/1 Points] DETAILS PREVIOUS ANSWERS OSCALC1 3.7.290. Use the given values to find (f )'(a). f(7) = 9, f'(7) = 2 =9 2 (F-1 ) ' (a) = 9 X T 10. [0/1 Points] DETAILS PREVIOUS ANSWERS OSCALC1 3.7.294. Use the given values to find (f-1)'(2). f(6) = 0, f'(6) = -9, a = 0 (F-1)' (a) = 6 X