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Given the curve :
`y= x^3 -x + 1`
`` We need to find the points on the graph such that the tangent lines parallel to the line y= 2x+5
First, we notice that the slope of the line y= 2x+ 5 is 2.
Then, the tangent lines should have slope of 2.
To find the slope of a tangent line, we need to find the derivative at the point of tendency.
Let us differentitae f(x).
`f'(x)= 3x^2 -1`
Then the slope is f'(a)= 2
`==> f'(a)= 3a^2 -1 = 2`
`==> 3a^2 = 3`
`==> a^2 = 1`
`a= +-1 `
Now we will find the values of f(x) at the tangent points.
`==> f(1)= 1^3 - 1 +1 = 1`
`f(-1)= -1^3 +1 +1 = 1```
Then, we have two points such that the tangent lines are parallel to the line y= 2x+5 .
Then, the points are: (1, 1) and (-1, 1).
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