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What is the equation of the line normal to #f(x)= -1/(5-x^2) # at #x=-3#?

Answer 1

Equation of line normal is #32x+12y+93=0#

Slope of the tangent of a curve, at a given point, is given by the value of the first derivative at that point. As normal is perpendicular to tangent, if #m# is the slope of tangent, slope of normal would be #-1/m#.

To find the point, let us put #x=-3# in #f(x)=-1/(5-x^2)# and #f(-3)=-1/(5-(-3)^2)=-1/-4=1/4#.

Hence, we are seeking normal at #(-3,1/4)#

#(df)/(dx)=-(-1/(5-x^2)^2)xx(-2x)=-(2x)/(5-x^2)^2#

Hence slope of tangent at #x=-3# is

#-(2(-3))/(5-(-3)^2)^2=6/(-4)^2=6/16=3/8#

and slope of normal is #-1/(3/8)=-8/3#

and equation of line normal is

#(y-1/4)=-8/3(x-(-3))#

or #12y-3=-32x-96# or #32x+12y+93=0#

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Answer 2

Paisley Johnson

The equation of the line normal to f(x) = -1/(5-x^2) at x = -3 is y = -3x - 4.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.