cal 1 at what points, if any does teh curve y=x^3-3x+4 have a horizontal tangent line?

Answer 1

if the first derivative of y ( with respect to x ) equals 0, say in point A
and the sign of the derivative is the not same on a small region around the point where dy(A)/dx = 0

Answer 2

y = x^3 - 3x + 4

To find all horizontal tangent lines, we want to find all tangent lines where the slope is equal to 0. We can find the slope of the tangent line at any given point of the graph by finding the first derivative and then making it 0.

dy/dx = 3x^2 - 3

Now, set dy/dx = 0,

0 = 3x^2 - 3
0 = 3(x^2 - 1)
0 = x^2 - 1
0 = (x - 1)(x + 1)

Therefore, x = {1, -1}

The curve has horizontal tangent lines at the points x = 1 and x = -1.

Answer 3

diferentiate to find the slope of the tangent then set equal to zero to find value of x where tangent is horizontal:

y=x^3 -3x +4
dy/dx = 3x^2 -3

3x^2 -3 =0
3x^2 = 3
x^2 = 1
x=+/-1 at x =1 y= 2 therefore the tangent is horizontal at the point (1,2) also
at x= -1 y =6 the second answer is (-1,6)

Answer 4

y=x^3-3x+4
y' = 3x^2-3 = 0
x = ±1

Therefore, the two horizontal tangent lines are,
y = 2 at x = 1
and
y = 6 at x = -1

Answer 5

Take the first derivative: y'
Set y' = 0
Solve for "x"
Take this "x" and solve for "y" (in the original equation).

This x,y will be the tangent point of the horizontal line.

Answer 6

Differentiate with respect to x. dy/dx = 3e^(3x) - e^x Let dy/dx = 0 and solve for x. 0 = 3e^(3x) - e^x Therefore, you have: x = -ln(3)/2 I hope this helps!

Answer 7

y' must be zero
y'=3x^2-3 =0 x=+-1
(1,2) and (-1,6)