How to Estimate Function Values Using Linearization 10:50 Ch 17. Saxon Calculus: Derivative as a... Ch 18. Saxon Calculus: Second... Ch 19. Saxon Calculus: Applications of the... Ch 20. Saxon Calculus: Computation of... Ch 21. Saxon Calculus: Riemann Sums Ch 22. Saxon Calculus: In...
When plugging −4x3(y2+1)+6x2y+2x(2y−3)−2=0 −4x3(y2+1)+6x2y+2x(2y−3)−2=0 into my calculator and trying to solve for x x i get x(2x2(y2+1)−3xy−2y+3)=−1 x(2x2(y2+1)−3xy−2y+3)=−1 as a result which i don't know what to with...
setting 3(x+sqrt(3))(x-sqrt(3))=0 gives x=-sqrt(3) and x=sqrt(3) from the second and third factors. The first factor, 3, doesn't give us a value. These values are the "x" values in the original function that are either local maximum or minimum...