Question #0671a

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Answer 1 · 2017-02-02T13:45:16

#(1,1) and (-1,1).#

Remember that the Slope of a tgt. line at the pt. #(x,y)# is #dy/dx.#

#because, y=x^3-x+1 rArr dy/dx=3x^2-1#.

The Slope of the given line is #2#, and, the tgt. line is parallel to this

line, the slope of tgt. line must also be #2.#

This gives, #3x^2-1=2 rArr 3x^2=3 :. x=+-1.#

Then, #y=x^3-x+1 rArr y=1" for "x=+-1.#

Accordingly, the desired points on the curve are #(1,1) and (-1,1).#

Answer 2 · 2017-02-02T14:02:35

The points on the curve #y=x^3-x+1# where the tangent line is parallel to the line #y=2x+5# are the points:

#P_1 = (-1,1)# with tangent line: #y=2x+3#

#P_2 = (1,1)# with tangent line: #y=2x-1#

If we use the slope-intercept form of the line equation, we know that two lines:

#y=m_1x+c_1#

#y=m_2x+c_2#

are parallel if #m_1=m_2#.

The general equation for the line tangent to #y=f(x)# in the point #(x_0,f(x_0))# is:

#y = f(x_0)+f'(x_0)(x-x_0)#

or, in slope-intercept form:

#y = f'(x_0)x+(f(x_0)-x_0f'(x_0) )#

This means that the points on the curve #y=x^3-x+1# where the tangent is parallel to the line #y=2x+5# are the points where:

#f'(x) = 2#

that is the roots of the equation:

#d/(dx) ( x^3-x+1 ) = 2#

#3x^2-1 =2#

#3x^2 = 3#

#x=+-1#

In conclusion the points where the tangent line to the curve is parallel to the line #y=2x+5# are the points:

#P_1 = (-1,1)# with tangent line: #y=2x+3#

#P_2 = (1,1)# with tangent line: #y=2x-1#