Question

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

Answer 1

The points on the curve are
`\left ( 0,6 \right )`,
`y=\left ( √(7),-43 \right )` and
`y=\left ( -√(7),-43 \right )`.

Explanation:

Step1

For horizontal tangent, derivative of the function and equate to zero.

Given

Equation of curve is given as follows:

`y=x^(4)-14x^(2)+6`

Step2

Calculation:

Derivate the function with respect to x and equate to zero as follows:

`y'=4x^(3)-28x+0=0`

`4x^(3)-28x+0=0`

`4x(x^(2)-7)=0`

Now,

`4x=0`

`X=0`

And,

`x^(2)-7=0`

`x=\pm √(7)`

Therefore, the values of
`x` are
`0, √(7) and -√(7)`.

Step3

Substitute the values of
`x` in the equation as follows:

For
`x=0`,

`Y=0-0+6`

`Y=0`

Therefore, the point is
`\left ( 0,6 \right )`.

For
`x=√(7)`,

`y=\left ( √(7) \right )^2-14\left ( √(7) \right )^2+6`

`y=49-14* 7+6`

`y=49-98+6`

`Y=-43`

Therefore, the point is
`y=\left ( √(7),-43 \right )`.

For
`x=-√(7)`,

`y=\left ( -√(7) \right )^4-14\left (- √(7) \right )^2+6`

`y=49-14* 7+6`

`y=49-98+6`

`Y=-43`

Therefore, the point is
`y=\left ( -√(7),-43 \right )`.