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10 Nov 2019
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Using the function y = f(x) obtained in (e), find the y-coordinate corresponding lo the value x = -1/2 Evaluate both formulas for dy/dx at this point. Why Jo you think this happens.? Using the explicit derivative formula obtained in pun (c), find all point on the graph of where the tangent line is horizontal. Then substitute the xĂÂ and y-coordinates of each such point into the implicit dy/dx formula given in part (b). Again, why docs this happen? Again using the explicit derivative formula from part (c). find all points on the graph of f where the tangent slope equals 6. Now substitute the x- and y- coordinates of each such point into the implicit formula given in part (b) to confirm this slope value. It is clear that no point (x, y) for which x + 2y = 0 will appear on the graph of original equation In effect, points on the line "cut out* of the graph of the function f. Relate this observation to the phenomena encountered earlier in parts (e) and (f). Once again referring to the explicit function f(x) obtained in part (c), where does the graph of f have a vertical asymptote? What happens when you substitute this X-value into the original implicit equation (given at the beginning of this worksheet) and attempt to solve toy ? Explain. [ Verification of the implicit derivative given in part (b).] Implicit) differentiate the equation applying the quotient role on the right-hand side. Clear the denominator and expand terms, then solve for and simplify.
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Using the function y = f(x) obtained in (e), find the y-coordinate corresponding lo the value x = -1/2 Evaluate both formulas for dy/dx at this point. Why Jo you think this happens.? Using the explicit derivative formula obtained in pun (c), find all point on the graph of where the tangent line is horizontal. Then substitute the xĂÂ and y-coordinates of each such point into the implicit dy/dx formula given in part (b). Again, why docs this happen? Again using the explicit derivative formula from part (c). find all points on the graph of f where the tangent slope equals 6. Now substitute the x- and y- coordinates of each such point into the implicit formula given in part (b) to confirm this slope value. It is clear that no point (x, y) for which x + 2y = 0 will appear on the graph of original equation In effect, points on the line "cut out* of the graph of the function f. Relate this observation to the phenomena encountered earlier in parts (e) and (f). Once again referring to the explicit function f(x) obtained in part (c), where does the graph of f have a vertical asymptote? What happens when you substitute this X-value into the original implicit equation (given at the beginning of this worksheet) and attempt to solve toy ? Explain. [ Verification of the implicit derivative given in part (b).] Implicit) differentiate the equation applying the quotient role on the right-hand side. Clear the denominator and expand terms, then solve for and simplify.
Keith LeannonLv2
1 Jun 2019