How can I find the points on the curve y = x^4 − 10x^2 + 1 where the tangent line is horizontal?

1 Answer
VNVDVI
Feb 26, 2018

Take the first derivative, set it equal to #0#, solve for #x#, plug back the solutions for #x# into the original function. The tangent line is then shown to be horizontal at #(0,1), (sqrt(5),-24), (-sqrt(5),-24).#

Explanation:

The tangent line is horizontal where the derivative is equal to zero.

Graphically interpreted, the derivative at a point on a curve is the slope at that point. We draw tangent lines at certain points to represent the slope at certain points. A horizontal line, #y=c# where #c# is a constant value, has a slope of zero, so wherever a horizontal tangent line exists, the derivative must be zero.

Take the first derivative with the Power Rule, set #y'=0#, and solve for #x#:

#y'=4x^3-(2)(10)x=4x^3-20x#

#4x^3-20x=0#

#4x(x^2-5)=0#

#x=0, +-sqrt(5)#

So, since #y'=0# at #x=0, x=+-sqrt(5)#, the tangent line is horizontal at those points. To find the #y#-values at these points, take #y(0), y(sqrt(5)),y(-sqrt(5)).#

#y(0)=0^4-10(0^2)+1=1#

The tangent line is horizontal at #(0,1).#

#y(sqrt(5))=sqrt(5)^4-10sqrt(5)^2+1=25-50+1=-24#

The tangent line is horizontal at #(sqrt(5),-24).#

#y(-sqrt(5))=(-sqrt(5)^4)-10(-sqrt(5)^2)+1=25-50+1=-24#

The tangent line is horizontal at #(-sqrt(5),-24).#

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