Question #6c32c

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Answer 1 · 2018-02-06T09:13:24

#x_1=1# and #x_2=1/3#

We want to find the horizontal tangent on the function

#f(x)=2x^3-4x^2+2x-3#

The graph must have a horizontal tangent when #f'(x)=0#

Differentiate the function by the power rule

#f'(x)=6x^2-8x+2#

Solve the quadratic equation #f'(x)=6x^2-8x+2=0#

#x=(8+-sqrt((-8^2)-4*6*2))/12#

#=(8+-4)/12#

#=(2+-1)/3#

The graph have a horizontal tangent at #x_1=1# and #x_2=1/3#