Calculus

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. 👍 0
  2. 👎 0
  3. 👁 940
  1. 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

    1. 👍 0
    2. 👎 0

Respond to this Question

First Name

Your Response

Similar Questions

  1. Calculus

    The curve y = x/(1 + x^2) is called a serpentine. Find an equation of the tangent line to this curve at the point(2, 0.40).(Round the slope and y-intercept to two decimal places.)

  2. calculus

    The point P(7, −4) lies on the curve y = 4/(6 − x). (a) If Q is the point (x, 4/(6 − x)), use your calculator to find the slope mPQ of the secant line PQ (correct to six decimal places) for the following values of x. (i) 6.9

  3. calculus

    1. Given the curve a. Find an expression for the slope of the curve at any point (x, y) on the curve. b. Write an equation for the line tangent to the curve at the point (2, 1) c. Find the coordinates of all other points on this

  4. Math

    Consider the graph of the parabola ​f(x)equals=x squaredx2. For xgreater than>0 and hgreater than>​0, the secant line through​ (x,f(x)) and ​(xplus+​h,f(x+h)) always has a greater slope than the tangent line at​

  1. last calc question, i promise!

    given the curve x + xy + 2y^2 = 6... a. find an expression for the slope of the curve. i got (-1-y)/(x + 4y) as my answer. b. write an equation for the line tangent to the curve at the point (2,1). i got y = (-1/3)x + (5/3). but i

  2. Calculus

    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 tangent to the curve has slope 1/2. (c) Show that there are now points (x,y) on the curve where the line

  3. Calculus

    The slope of the tangent line to a curve at any point (x, y) on the curve is x divided by y. What is the equation of the curve if (2, 1) is a point on the curve?

  4. Calculus

    A) The point p(3,1) lies on the curve y= square root x-2. If Q is the point (x, square root x-2) use your calculator to find the slope of the decant line PQ for the following values of x 1) 2.5 2.) 2.9 3.) 2.99 4.) 2.999 5.) 3.5

  1. Calculus

    At what point on the curve x = 3t^2 + 2, y=t^3-8 does the tangent line have slope 1/2?

  2. math

    For the curve given by 4x^2+y^2=48+2xy, show that there is a point P with x-coordinate 2 at which the line tangent to the curve at P is horizontal.

  3. calculus 1

    The point P(2,-1) lies on the curve y=1/(1-x) If Q is the point (x, 1/(1-x) find slope of secant line. these are the points 2, -1 1.5,2 1.9,1.111111 1.99,1.010101 1.999,001001 2.5,0.666667 2.1,0.909091 2.01,0.990099 2.001,0.999001

  4. calculus

    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 horizontal tangent line to the curve. c. The line through the origin with slope -1 is tangent to the

You can view more similar questions or ask a new question.