Solution for 1 y = Va? Here's an outline y = x^4ax^3bx^2cxd y' = 4x^33ax^22bxc Knowing the equations of the tangent lines at x=0 and x = 1 allows us to find y and y' at those values of x At x=0, we get y=2(0)1=1 and y'=2 At x=1, we get y=23(1)=1 and y'=3 Using the formulas above we get 4 equations with 4 unknowns;Divide \frac{3}{y1}, the coefficient of the x term, by 2 to get \frac{3}{2\left(y1\right)} Then add the square of \frac{3}{2\left(y1\right)} to both sides of the equation This step makes the left hand side of the equation a perfect square
Solved Al Consider The Region R Bounded By Y X2 2x 2 And Y 2x 1 A Sketch R And Label Any Relevant Points B Set Up A Defin Course Hero
