At which x values does the curve y=1+60x^3-10x^5 does the tangent line have the largest slope. List the x values in increasing order and enter n in the second box if the max occurs at only one x value
Well, that's a pretty steep question! Let's put on our math hats and try to tackle it.
To find the x-values where the tangent line has the largest slope, we need to find where the derivative of the function is equal to zero. We can start by finding the derivative of the function y=1+60x^3-10x^5.
The derivative of y with respect to x is:
y' = 180x^2 - 50x^4
Now, we can set y' equal to zero and solve for x:
180x^2 - 50x^4 = 0
Factoring out an x^2:
x^2(180 - 50x^2) = 0
Now, we have two possibilities:
1. x^2 = 0
Solving this equation, we find x = 0.
2. 180 - 50x^2 = 0
Rearranging the equation:
50x^2 = 180
Dividing by 50:
x^2 = 3.6
Taking the square root:
x = √3.6
So, we have two x-values where the tangent line could have the largest slope: 0 and √3.6.
Listing them in increasing order, we have: 0, √3.6
So, the answer is 0, √3.6. There are two x-values where the tangent line has the largest slope. No need to enter 'n' in the second box. Hope this helps, and remember, math can be a real roller coaster ride!
To find the x values at which the tangent line to the curve y=1+60x^3-10x^5 has the largest slope, we need to find the values of x where the derivative of the function is equal to zero.
Given that y = 1 + 60x^3 - 10x^5, let's find its derivative with respect to x:
dy/dx = 0 + 180x^2 - 50x^4.
To find the critical points, we set the derivative equal to zero and solve for x:
180x^2 - 50x^4 = 0.
Factoring out x^2, we get:
x^2(180 - 50x^2) = 0.
From this equation, we can see that there are two possible solutions for x: x = 0 and x = ±√(180/50) ≈ ±2.6833.
Therefore, the x values at which the tangent line has the largest slope are x = -2.6833, 0, and 2.6833.
Listing them in increasing order, we have: -2.6833, 0, 2.6833.
Therefore, the correct answer is "-2.6833, 0, 2.6833".
To find the x values at which the tangent line has the largest slope for the given curve, we need to find the critical points of the curve.
First, let's find the derivative of the function y = 1 + 60x^3 - 10x^5 with respect to x. The derivative will give us the slope of the tangent line at any given point.
So, differentiating the function, we have:
dy/dx = 0 + 180x^2 - 50x^4
To find the critical points, we set the derivative equal to zero and solve for x:
180x^2 - 50x^4 = 0
Now, let's factor the equation:
x^2(180 - 50x^2) = 0
Setting each factor equal to zero, we get two possibilities:
x^2 = 0 --> x = 0
180 - 50x^2 = 0 --> 50x^2 = 180 --> x^2 = 180/50 --> x^2 = 3.6
Taking the square root of both sides, we have:
x = ±√(3.6)
Calculating the square root, we get:
x ≈ ±1.897
So, the critical points are x = -1.897, 0, and 1.897.
To determine where the tangent line has the largest slope, we need to find the maximum value of the derivative. We can do this by checking for any changes in sign in the derivative.
Evaluating the derivative at x = -1.897, we get:
dy/dx ≈ 0 + 180(-1.897)^2 - 50(-1.897)^4 ≈ 685.161
Evaluating the derivative at x = 0, we get:
dy/dx ≈ 0 + 180(0)^2 - 50(0)^4 ≈ 0
Evaluating the derivative at x = 1.897, we get:
dy/dx ≈ 0 + 180(1.897)^2 - 50(1.897)^4 ≈ -685.161
Since the derivative changes sign from positive to negative as x moves from -1.897 to 1.897, the maximum slope occurs at those x values.
Therefore, the x values at which the tangent line has the largest slope are -1.897 and 1.897, in increasing order.
So, the answer is: -1.897, 1.897
the slope is given by dy/dx
dy/dx = 180x^2 - 50x^4
This will have a max (or min) if its derivative is zero, that is,
360x - 200x^3 = 0
x(360 - 200x^2) = 0
x = 0 or x = ±360/200 = ±9/5
when x = 0, dy/dx = 180(0) - 50(0) = 0
when x = 9/5 , dy/dx = 180(81/25) - 50(6561/625)
= 1458/25
when x = -9/5 , dy/dx = 180(-81/25) - 50(-6561/625)
= -1458/25
So,when x = -9/5 the slope is a minimum
when x = 9/5 the slope is a maximum
check my arithmetic.