Ask a New Question
Menu

Show that the curve y=6x^3+5x-3 has no tangent line with slope 4.

The answer key says that

m=y'=18x^2+5, but x^2 is greater than or equal to 0 for all x, so m is greater than or equal to 5 for all x.

I don't understand that x^2 is greater than or equal to zero. Where did that come from? Why isn't x all real numbers?

1 answer

y' = 18x^2 + 5

we want 18x^2 + 5 = 4
18x^2 = -1
x^2 = -1/18
but that is not possible since the square of anything has to be positive,
so there is value of x making this equations true
thus there is not slope of 4
thuse there is no such tangent
Similar Questions
  1. Consider the curve defined by 2y^3+6X^2(y)- 12x^2 +6y=1 .a. Show that dy/dx= (4x-2xy)/(x^2+y^2+1) b. Write an equation of each
    1. answers icon 3 answers
  2. Consider the curve given by y^2 = 4xy + 1. Show work for all parts.A. Find dy/dx. B. Find all points on the curve where the
    1. answers icon 1 answer
  3. Consider the curve given by y^2 = 2+xy(a) show that dy/dx= y/(2y-x) (b) Find all points (x,y) on the curve where the line
    1. answers icon 2 answers
  4. original curve: 2y^3+6(x^2)y-12x^2+6y=1dy/dx=(4x-2xy)/(x^2+y^2+1) a) write an equation of each horizontal tangent line to the
    1. answers icon 1 answer
more similar questions