Find 2 x values for normal which have the gradient -1/12?

equation of the curve = y = x^3 + 3x - 1

find two x values for which the normal to the curve have a gradient of -1/12?

1 Answer
Mar 18, 2018

#color(blue)(x=sqrt(3)\ \ \ \)# and #\ \ \ \color(blue)(x=-sqrt(3))#

Explanation:

If #bbm_1# is the gradient of the normal to a curve at a particular point, and #bbm_2# is the gradient of the tangent at the same point, then:

#m_1*m_2=-1#

Let #m_1# be the gradient of the normal, given:

#m_1=-1/12#

If #m_2# be the gradient of the tangent then:

#m_2=1/(m_1)=-1/(-1/12)=12#

We can find the #bbx# values at the point or points, by using the first derivative, and equating it to #bb12#. Since this function gives us the gradient at a point and we already know this has to be #bb12#

Differentiating

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

Equating to 12:

#3x^2+3=12#

#3x^2=9#

#x^2=3=>x=+-sqrt(3)#

Putting these values in #f(x)#, we can obtain two corresponding #y# values:

#f(sqrt(3))=(sqrt(3))^3+3(sqrt(3))-1=6sqrt(3)-1#

#f(-sqrt(3))=(-sqrt(3))^3+3(-sqrt(3))-1=-6sqrt(3)-1#

Using point slope form of a line:

#(y_2-y_1)=m(x_2-x_1)#

For tangent lines we have #m=12#

#y-(6sqrt(3)-1)=12(x-(sqrt(3))#

#color(blue)(y=12x-6sqrt(3)-1\ \ \ \ \ \ \ \ \ \)# tangent line 1

#y-(-6sqrt(3)-1)=12(x-(-sqrt(3))#

#color(blue)(y=12x+6sqrt(3)-1\ \ \ \ \ \ \ \ \ \)# tangent line 2

For normals we have #bb(m=-1/12)#

As above:

#y-(6sqrt(3)-1)=-1/12(x-(sqrt(3))#

#color(blue)(y=-1/12x+73/12sqrt(3)-1\ \ \ \ \ \ \ \)# normal line 1

#y-(-6sqrt(3)-1)=-1/12(x-(-sqrt(3))#

#color(blue)(y=-1/12x-73/12sqrt(3)-1\ \ \ \ \ \ \ \)# normal line 2

GRAPH:

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