Respuesta :
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].
