Find the equation of the tangent line

Question

Let f(x) = (x⁴ + x³ + x² + x + 1)(x⁵ + x³ + x + 2).

Find the equation of the tangent line to the curve y = f(x) at the point on the curve whose x-coordinate is −1.

 

Book answer: y=10x + 11. Mine does not match and I can't figure out why.

Answer 1

First, evaluate f(1) to find the point of tangency.

 

f(-1) = (1 - 1 + 1 - 1 + 1) (-1 - 1 - 1 + 2)

       = 1 * -1

       = -1

 

The point of tangency is (-1, -1).

 

Then, find the derivative of f(x) and evaluate it at x=-1.  Call this f'(-1), which is the slope of the tangent line.

To find the derivative of f(x), use product rule.

 

I will let you find f'(-1) on your own.

 

 

Then, use the point-slope form of a line to find the tangent line.

 

y + 1 = f'(-1)[x + 1]

Answer 2

First, you can find f(-1) =(1+-1+1+-1+1)(-1+-1+-1+2) = (1)(-1) = -1 So your tangent line must pass through (-1,-1).

 

Just using this, the answer your book has an incorrect answer because (-1,-1) does not satisfy y=10x+11 and is therefore not on the line....unless you've copied the problem wrong.

 

Next, you must find the derivative so that you can plug in -1 and find the slope of the tangent at -1. So we must use the product rule which gives us

 

f'(x) = (x⁴ + x³ + x² + x + 1)(5x⁴ + 3x² + 1) + ( 4x³ + 3x² + 2x + 1)(x⁵+ x³ +x+2)

f'(-1) = ( 1 + -1 + 1 + -1 + 1)(5+3+1) + (-4+3+-2+1)(-1 + -1 + -1 + 2)

        = (1)(9) + (-2)(-1)

        = 11

 

So I got that the point is (-1,-1) and the slope is 11, so the equation is

y+1 = 11(x+1)

y    = 11x+10.

 

hope this helped.