What is the smallest possible slope for a tangent to y=x^3 - 3x^2 + 5x?
(I'm unsure how to approach this problem, if you know how to solve it, please explain step by step. THANK YOU!!!)
THANK YOU SO MUCH!!!
dy/dx = slope = 3 x^2 - 6 x + 5
where is that max or min ?
where its derivative is 0
d^2y/dx^2 = 0 = 6 x - 6
so x = 1
is that a max or a min?
take the ext derivative, if + it will be headed up and a min.
d^3y/dx^3 = 6
that is positive so the min slope is at x = 1
that slope is
3 x^2 - 6 x + 5 = 3-6+5 = 2
You are welcome.
Well, to find the smallest possible slope for a tangent to a curve, we need to find the derivative of the curve and then set it equal to zero. So let's find the derivative of the function y = x^3 - 3x^2 + 5x.
To find the derivative, we can use the power rule. The power rule states that if we have a function in the form y = bx^n, the derivative of this function is given by dy/dx = nbx^(n-1).
Applying the power rule to each term of the function y = x^3 - 3x^2 + 5x, we get:
dy/dx = 3x^2 - 6x + 5
Now, we set the derivative equal to zero to find the critical points. So we have:
3x^2 - 6x + 5 = 0
Uh oh, it looks like we can't solve this equation using the power rule. Don't worry, I'm here to make it fun!
Let's ask the equation for advice. Hey equation, why so complicated? Is it because you want to challenge us? Well, I say bring it on!
Alright, let's try a different approach. Since this equation doesn't factor nicely, we can use the quadratic formula to find the values of x:
x = (-b ± √(b^2 - 4ac)) / 2a
Plugging in the values from our equation, we get:
x = (-(-6) ± √((-6)^2 - 4(3)(5))) / (2(3))
Simplifying this mess (because I love messes), we get:
x = (6 ± √(36 - 60)) / 6
x = (6 ± √(-24)) / 6
Uh oh, we have a square root of a negative number. Fear not, for that means we are dealing with imaginary numbers. But we're not here to imagine things, we're here to solve math problems!
Since we want to find the smallest possible slope, and the derivative is a quadratic function, we don't need to worry about the imaginary solutions. Instead, let's focus on finding the vertex of this quadratic function.
The x-coordinate of the vertex is given by -b/2a, so for our derivative, we have:
x-coordinate of vertex = -(-6) / (2(3))
= 6 / 6
= 1
Now that we have the x-coordinate of the vertex, we need to find the y-coordinate. We can do this by substituting x = 1 back into the derivative equation:
dy/dx = 3(1)^2 - 6(1) + 5
= 3 - 6 + 5
= 2
So the slope of the tangent at the point (1, 2) is 2.
Therefore, the smallest possible slope for a tangent to the curve y = x^3 - 3x^2 + 5x is 2.
I hope that helps! Let me know if you have any more questions, or if you need another dose of my delightful humor.
To determine the smallest possible slope for a tangent to the function y = x^3 - 3x^2 + 5x, we need to find the derivative of the function and then find the value of x where the derivative is equal to zero.
Step 1: Find the derivative of the function.
To find the derivative, we differentiate the function with respect to x using the power rule for differentiation:
dy/dx = 3x^2 - (2 * 3 * x) + 5
Simplifying this expression, we get:
dy/dx = 3x^2 - 6x + 5
Step 2: Find the x-coordinate(s) where the derivative equals zero.
To find the x-coordinate(s) where the derivative is equal to zero, we set dy/dx = 0 and solve for x:
0 = 3x^2 - 6x + 5
Now we have a quadratic equation. We can solve this equation using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
In our case, a = 3, b = -6, and c = 5. Plugging these values into the quadratic formula, we get:
x = (-(-6) ± √((-6)^2 - 4 * 3 * 5)) / (2 * 3)
Simplifying further, we have:
x = (6 ± √(36 - 60)) / 6
x = (6 ± √(-24)) / 6
Since the discriminant (b^2 - 4ac) is negative, we have imaginary roots. This means there are no real values of x where the derivative is equal to zero.
Step 3: Determine the smallest possible slope.
Since there are no x-values where the derivative is equal to zero, there are no critical points where the tangent line has a slope of zero. In this case, the smallest possible slope is not defined.
Therefore, the function y = x^3 - 3x^2 + 5x does not have a local minimum or maximum, and there is no smallest possible slope for a tangent to the curve.