Find all points on the curve y=x^3-x+1 where the tangent line is parallel to the line y=2x+5

Sunday, November 8, 2015

Given the curve :

$\displaystyle{y}={{x}}^{{3}}-{x}$
+ 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).

$\displaystyle{f{'}}{\left({x}\right)}={3}{{x}}^{{2}}-{1}$

Let
a be the point of tendency such that the tangent line at the point x=a, is parallel to
the line 2x+5.

Then the slope is f'(a)=
2

$\displaystyle=\Rightarrow{f{'}}{\left({a}\right)}={3}{{a}}^{{2}}-{1}=$
2

$\displaystyle=\Rightarrow{3}{{a}}^{{2}}=$
3

$\displaystyle=\Rightarrow{{a}}^{{2}}={1}$

$\displaystyle{a}=$
+-1

Now we will find the values of f(x) at the tangent
points.

$\displaystyle=\Rightarrow{f{{\left({1}\right)}}}={{1}}^{{3}}-{1}+{1}=$
1

$\displaystyle{f{{\left(-{1}\right)}}}=-{{1}}^{{3}}+{1}+{1}=$
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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