Find the points on the curve y = x4 − 14x2 + 6 where the tangent line is horizontal.

Answer:The points on the curve are [tex]\left ( 0,6 \right )[/tex], [tex]y=\left ( \sqrt{7},-43 \right )[/tex] and [tex]y=\left ( -\sqrt{7},-43 \right )[/tex]. Step-by-step explanation:Step1 For horizontal tangent, derivative of the function and equate to zero. Given Equation of curve is given as follows: [tex]y=x^{4}-14x^{2}+6[/tex]Step2 Calculation: Derivate the function with respect to x and equate to zero as follows: [tex]y'=4x^{3}-28x+0=0[/tex][tex]4x^{3}-28x+0=0[/tex][tex]4x(x^{2}-7)=0[/tex] Now, [tex]4x=0[/tex][tex]X=0[/tex]And, [tex]x^{2}-7=0[/tex][tex]x=\pm \sqrt{7}[/tex]Therefore, the values of [tex]x[/tex] are [tex]0, \sqrt{7} and -\sqrt{7}[/tex].Step3 Substitute the values of [tex]x[/tex] in the equation as follows: For [tex]x=0[/tex], [tex]Y=0-0+6[/tex] [tex]Y=0[/tex]Therefore, the point is [tex]\left ( 0,6 \right )[/tex].For [tex]x=\sqrt{7}[/tex], [tex]y=\left ( \sqrt{7} \right )^2-14\left ( \sqrt{7} \right )^2+6[/tex][tex]y=49-14\times 7+6[/tex][tex]y=49-98+6[/tex][tex]Y=-43[/tex]Therefore, the point is [tex]y=\left ( \sqrt{7},-43 \right )[/tex]. For [tex]x=-\sqrt{7}[/tex],[tex]y=\left ( -\sqrt{7} \right )^4-14\left (- \sqrt{7} \right )^2+6[/tex][tex]y=49-14\times 7+6[/tex][tex]y=49-98+6[/tex][tex]Y=-43[/tex]Therefore, the point is [tex]y=\left ( -\sqrt{7},-43 \right )[/tex].