Question #0671a

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Answer 1 · 2017-02-02T13:45:16

$(1, 1) and (- 1, 1) .$ $(1,1) and (-1,1).$

Remember that the Slope of a tgt. line at the pt. $(x, y)$ $(x,y)$ is $\frac{d y}{d x} .$ $dy/dx.$

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

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: $y=2x-1$