Question

Let f be the function defined by f(x) =x^4 -3x^2 +2

A)find the zeros of f
B)write an equation of the line tangent to the graph of f at the point where x=1
C) find the x coordinate of each point at which the line tangent to the graph of f is parallel to the line y=-2x+4

Answer 1

The zeros of the function can be determined by equating the equation to zero and determining the values of x.
0 = x4 - 3x2 + 2 the roots are x1 = sqrt 2 x2 = -sqrt 2x3 = 1x4 = -1
The tangent line is determined by differentiating the polynomial and substituting x by 1 to get the slope. That is, m = 4x^3 - 6 x = 4*1 - 6 = -2y-y1 = m(x-x1)when x =1 , y = 1-3+2 = 0
y - 0 = -2*(x-1)y = -2x + 2
c. y = -2x + 4 m = -2 -2 = 4x3 - 6x x= -1.3666 ; y =-0.1160 x= 1.3666 ; y =--0.1160
x =1 ; y = 0

Answer 2

A convenient way to find the zeros of
`f(x)=x^4-3x^2 +2` is by factoring.

a) The equation,

`x^4-3x^2 +2=0`

can be rewritten as,

`(x^2)^2-3(x^2) +2=0`

We can think of this equation as a quadratic equation in
`x^2`, with
`a=1,b=-3\:\: and \:\: c=2`.

Observe that
`ac=1 * 2=2`.

We find two factors of 2 that adds up to
`-3`. These are,
`-2,-1`.

Now let us split the middle term. to obtain;

`(x^2)^2-(x^2) -2(x^2)+2=0`

We can factor to get,

`(x^2)(x^2-1)-2((x^2-1)=0`

We factor further to obtain;

`(x^2-1)((x^2-2)=0`

`\Rightarrow (x-1)(x+1((x- √(2))(x+ √(2))=0`

Hence the zeroes of
`f(x)` are;

`x=1,x=-1,x= √(2),x=- √(2)`

b) To find the line tangent, we must first, find the slope using differentiation. That is,

`Slope\:\: function=f'(x)=4x^3-6x`

At
`x=1`,

`Slope=f'(1)=4(1)^3-6(1)=-2`

Also, we need to determine the
`y` value at
`x=1`. That is;

`f(1)=(1)^4-3(1)^2+2=0`

Now we can use the slope
`m=-2` and the point
`(1,0)` to write ythe equation of the line tangent.

`y-y_1 =m(x-x_1)`

`\Rightarrow y-0 =-2(x-1)`

`\Rightarrow y=-2x+2`

c)

If the line tangent is parallel to the line
`y=-2x+4`, then

`f'(x)=-2`

Since parallel lines have the same slope.

`\Righarrow 4x^3-6x=-2`

`\Righarrow 4x^3-6x+2=0`

`\Righarrow (x-1)(2x- ( √(3)-1)(2x+ ( √(3)-1)=0`

Hence the x-coordinates are,

`x=1,x= ( √(3)-1)/(2),x= -(√(3)-1)/(2)`