Find 2 tangent line equations to the curve y=x^3+x at the points where the slope of the curve is 4.
What is the smallest possible slope of the curve?
At what x-value(s)does the curve have this slope?
So far I have figured out the derivative of this function is
y'= 3x^2+1.
I think one of the equations is y-2 = 4(x-1).
I think smallest slope is 4.
I think x=1 and x=-1.
for the two tangent lines, where is 3x^2+1 = 4?
x = ±1
y(1)=2
y(-1) = -2
So, the two lines are
y-2 = 4(x-1)
y+2 = 4(x+1)
y" = 6x
so, at x=0 the slope is smallest: 1
If you look at the graph, you can clearly see that the slope at x=0 is smaller than at x=1 or -1.
http://www.wolframalpha.com/input/?i=x%5E3%2Bx
To find the equations of the tangent lines to the curve y = x^3 + x at the points where the slope of the curve is 4, you need to follow a few steps:
Step 1: Find the derivative of the function.
You correctly found the derivative of y = x^3 + x as y' = 3x^2 + 1.
Step 2: Set the derivative equal to the given slope of 4.
Since the slope of the tangent line is 4, set y' = 4 and solve for x:
3x^2 + 1 = 4
Step 3: Solve the equation for x.
Subtract 1 from both sides of the equation:
3x^2 = 3
Divide both sides of the equation by 3:
x^2 = 1
Take the square root of both sides to solve for x:
x = ±1
Step 4: Find the corresponding y-values.
Substitute the values of x into the original function y = x^3 + x:
For x = 1, y = 1^3 + 1 = 2
For x = -1, y = (-1)^3 + (-1) = -2
Step 5: Write the equation of the tangent lines.
The equation of a line in point-slope form is y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope.
For the point (1, 2) and slope 4:
y - 2 = 4(x - 1)
Simplifying, you get:
y - 2 = 4x - 4
And for the point (-1, -2) and slope 4:
y - (-2) = 4(x - (-1))
Simplifying, you get:
y + 2 = 4x + 4
So, the equations of the tangent lines to the curve y = x^3 + x at the points where the slope is 4 are:
y - 2 = 4x - 4
y + 2 = 4x + 4
Moving on to the next part of the question, the smallest possible slope of the curve can be found by finding the minimum value of the derivative. Looking at y' = 3x^2 + 1, the derivative is a positive quadratic function, so it does not have a minimum value. Therefore, there is no smallest possible slope.
However, if you are referring to the values of x where the slope is the closest to 0, you can set the derivative equal to 0 and solve for x:
3x^2 + 1 = 0
3x^2 = -1
x^2 = -1/3
Since the square of a real number cannot be negative, there are no real x-values that satisfy this equation. Therefore, there are no x-values where the slope of the curve is the closest to 0.