If #f(x)= x^4-4x^3+4x^2-1#, how do you find all values of #x# where the graph of #f# has a horizontal tangent line?

Question

Charles Arocha · Accepted Answer

Take the derivative of #f(x)# and set it equal to #0# to find #f(x)# has horizontal tangents at #x=0,1,2#.

A horizontal tangent line occurs when the slope is #0#. It's just a line - it's no changing, so the slope must be #0#. Therefore, to find the #x#-values for which #f(x)# has a horizontal tangent line, we take the derivative of #f(x)#, set it equal to #0#, and solve. Beginning with the derivative: #f'(x)=4x^3-12x^2+8x# Setting it equal to #0# and solving: #0=4x^3-12x^2+8x# #0=x(4x^2-12x+8)# #0=x(2x-2)(2x-4)# #x=0# #2x-2=0->x=1# #2x-4=0->x=2# Therefore #f(x)# has horizontal tangents at #x=0#, #x=1#, and #x=2#. We can confirm this from the graph of #f(x)#: graph{x^4-4x^3+4x^2-1 [-10, 10, -5, 5]} You can see that at all of these #x#-values, the graph is neither increasing nor decreasing, which means the slope of the tangent line is #0# (and thus it is a horizontal tangent).

Ezekiel Alexander · Answer

undefined To find all values of $ x $ where the graph of $ f(x) = x^4 - 4x^3 + 4x^2 - 1 $ has a horizontal tangent line, you need to find where the derivative of $ f(x) $ is equal to zero. So, you need to find $ f'(x) $ and solve the equation $ f'(x) = 0 $ for $ x $. Then, you can determine the corresponding values of $ x $.