Home > Calculus > First Principles Example 1: x²

For what values of x is dy/dx zero and undefined?

#y = "x^2 - 3x + 1"/"x+2"#

Answer 1

#dy/dx# is zero for #x =-2 pm sqrt(11)#, and #dy/dx# is undefined for #x=-2#

Find the derivative: #dy/dx = (d(x^2 - 3x + 1))/dx 1/(x+2) + (x^2 - 3x + 1)(d)/(dx)(1/(x + 2))# #=(2x-3)/(x+2) - (x^2 - 3x + 1)1/(x + 2)^2# #=((2x-3)(x+2) - (x^2 - 3x + 1))/(x + 2)^2# #= (2x^2 - 3x + 4x -6 - x^2+3x-1)/(x + 2)^2# #= (x^2 + 4x -7)/(x + 2)^2# by the product rule and various simplifications.

Find zeros: #dy/dx=0# if and only if #x^2 + 4x -7=0#. The roots of this polynomial are #x_{1,2} = (1/2)(-4 pm sqrt(4^2 - 4(-7))) = -2 pm sqrt(11)#, so #dy/dx=0# for #x = -2 pm sqrt(11)#.

Find where #dy/dx# is undefined: Since division by #0# is not allowed, #dy/dx# is undefined where #(x + 2)^2=0#, that is, where #x=-2#.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.