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In Problems 26 through 31 find the equation of the line that is tangent of the given function at the specified point. y = x5 - 3x3 - 5x - 2; (1, -5) y = -x3 - 5x2 + 3x - 1; (-1, -8) y = -x2 + 16/x2; (4, -7) y = 1 - 1/x + 2/ ; (4, 7/4) y = 2x4 - + 3/x; (1,4) y = (x2 - x)(3 - 2x); (-1, 2) Show transcribed image text
i need answers.
In Problems 26 through 31 find the equation of the line that is tangent of the given function at the specified point. y = x5 - 3x3 - 5x - 2; (1, -5) y = -x3 - 5x2 + 3x - 1; (-1, -8) y = -x2 + 16/x2; (4, -7) y = 1 - 1/x + 2/ ; (4, 7/4) y = 2x4 - + 3/x; (1,4) y = (x2 - x)(3 - 2x); (-1, 2)
Show transcribed image text Lelia LubowitzLv2
22 Oct 2019