Expert: Josh - 3/30/2010

Question
Hi Josh,
Struggling again!
Can you help me please with a number of questions.

For each function determine the equation of the tangent at points
1)         F(x) = xe^x      at x=0
2)         f(x)  =sin 2x     at x=0   and pi/6
Also
Finding the equation of each tangent of the function when
F(x) = x^3 -5x^3 +5x -4    which is parallel to the line   y=2x +1

and finding the equation of each tangent of the function
F(x) = x^3 +x^2 +x +1 which is perpendicular to the line 2y+x+5=0

And
For the function given determine the equations of the tangent and normal at each point indicated
F(x) = 2x^3 -5x +4       at x= -1  and      1

And finally
Find the x co-ordinate of the point where the normal to        F(x) = x^2 -3x +1          at x= -1 intersects the curve again

Thanks for your time,
Richard.

Answer
Hi Richard,

Life is a struggle. It's okay to ask for help.

1) Consider a curve with a shape that resembles a smoothed version of the letter "W" (we remove all the sharp corners). In this case, we cannot specify a tangent right in the middle (at x=0) because the function has higher values both to its left and to its right. If we draw a horizontal line at the peak around x=0, this will certainly intersect with the function at 2 points and defeats the purpose of a tangent which by definition only touches the function at a single point.

We don't face this problem with f(x)=x*e^x because f(x) is a strictly decreasing function [i.e., value of f(x) decreases as x increases] for x<-1; and f(x) is a strictly increasing function for x>-1.
This means, in the interval of interest (behavior around x=0), the value of f(x) increases as we increase x. This "monotonic increasing" property allows us to always draw a line tangential to f(x)=x*e^x (without crossing f(x) again) for positive values of x. In fact, this is true anywhere right of the stationary point x=-1.

Taking the 1st derivative to find a curve that only touches the function, f'(x)=x*e^x+1*e^x using the product rule of differentiation. So, the slope is given by f'(x)=(x+1)*e^x. Evaluating at x=0, slope "m"=(0+1)*e^0=1. So, the tangent at x=0 is actually y=x+b. Now, let's find "b".

Substituting x=0 into f(x) gives a point (x,y)=(0,0) that lies on both the function and the tangent. So, (0,0) has to satisfy the equation y=x+b for the tangent line, this gives b=0.

2) For f(x)=sin(2x), df/dx = 2*cos(2x).
  - Find slope m(x)=2*cos(2x) at x=0 and x=pi/6.
    You know that tangent has the form y=m*x+b.
  - To find b, first compute the (x,y) coordinates for the point common
    to both f(x) and the tangent, at x=0 and x=pi/6, respectively.
    Substitute said (x,y) values into y=m*x+b to get "b".

3) Do you mean "F(x) = x^3 -5x^2 +5x -4"? (x square in the 2nd term??)
  - Slope for tangent "m" is given by the first derivative F'(x)=3x^2-10x^2+5.
  - If m is parallel to y=2x+1, then m=2.
  Setting F'(x)=2, gives a quadratic equation 3x^2-10x^2+3=0.
  i.e., (3x-1)(x-3)=0
  i.e., x=1/3 and x=3 (check this yourself to ensure this is correct)
  This constraints the locations where a tangent can have a slope of 2 (parallel to the line).
  - The general form of the tangent equation is y=mx+b.
  - Already, we know m=2, as we want it to be parallel to y=2x+1.
  - It remains to find b.
    Here, we substitute x0=1/3 into F(x) to find y0. Plug x=x0, y=y0 into y=mx+b0 to find b0.
    Similarly, substitute x1=3 into F(x) to find y1. Plug x=x0, y=y0 into y=mx+b1 to find b1.

4) Tangent for F(x) = x^3 +x^2 +x +1 perpendicular to 2y+x+5=0.
  - Note: 2y+x+5=0 is equivalent to y=(-x-5)/2. Line has slope n=-1/2.
  - Use exactly the same strategy as Q3.
  - Only difference is that slope of tangent (we still call this "m")
    is perpendicular to the given line.
    The condition m*n=-1 must be satisfied given two slope "m" and "n" for two
    perpendicular lines. So, the required slope m = -1/n = 2.

5) Tangent has slope m given by dF/dx=6x-5.
  Normal has slope n given by -1/m = 1/(5-6x).
  Next, let x0=-1 and x1=+1, find the y-coordinates (y0 and y1) of the point
  common to both the function F(x) and the tangent (or normal, whatever the case may be).
  > To make clear, plug x=-1 into y=2x^3-5x+4 to find y0.
  >                plug x=+1 into y=2x^3-5x+4 to find y1.
  Equation of tangent 0: y=mx+b, where m equals dF/dx evaluated at x=x0.
                         value of b is found by substituting (x0,y0) into y=mx+b.
  Equation of normal 0: y=nx+b, see expression for "n" from above, evaluate it at x=x0.
                         value of b is found by substituting (x0,y0) into y=nx+b
  Equation of tangent 1: y=mx+b, where m equals dF/dx evaluated at x=x1.
                         value of b is found by substituting (x1,y1) into y=mx+b.
  Equation of normal 1: y=nx+b, see expression for "n" from above, evaluate it at x=x1.
                         value of b is found by substituting (x1,y1) into y=nx+b

6) Normal to F(x) has slope n = -1/(dF/dx), where dF/dx=2x-3.
  At x=-1, n=-1/(-2-3)=1/5.
  y=n*x+b  i.e., y=(x/5)+b
  As before, we need to find a point (x0,y0) common to both F(x) and normal, in order to find "b".

  If this intersects F(x) again, we equate F(x) with y.
  This gives x^2-3x+1 = (x/5)+b. You would have already found "b" at this point.
  So, just solve for x.

Yeah, I've focused more on outlining the approach this time.
I may have made mistakes along the way. I hope I managed to get the point across.
I'll let you sort out the details yourself.
Let me know if you have any problems.
 
Cheers,
Josh