How do you determine at which the graph of the function #y=1/x^2# has a horizontal tangent line?

Question

Cameron Alexander · Accepted Answer

By using derivatives

derivatives define the slope of a tangent line at a point on the function therefore if the tangent line is horizontal, its slope is 0 so, on differentiating #y'(x) = d/dx 1/x^2= 0# we're setting it equal to zero because we want to see the points at which the derivative is 0 so its slope Is 0 we can use the power rule here as #1/x^2# is just #x^-2# therefore, #-2x^-3 = 0# divide both sides by #-2# #= 1/x^3 = 0# this is an indeterminate form as the only way this could satisfy the equation is if x was positive or negative infinity therefore at finite values of x, we don't ever have a point where the tangent lines are horizontal you can see this on the graph that as x becomes bigger and bigger its slope decreases and gets closer and closer to 0, so as x approaches infinity, its slope approaches 0 graph{1/x^2 [-12.66, 12.65, -6.33, 6.33]}

Aurora Alvarado · Answer

undefined To determine where the graph of the function y=1/x^2 has a horizontal tangent line, we need to find the points where the derivative of the function is equal to zero. Taking the derivative of y=1/x^2 using the power rule, we get dy/dx = -2/x^3. Setting this derivative equal to zero and solving for x, we find that x=0. Therefore, the graph of the function y=1/x^2 has a horizontal tangent line at x=0.