At what value(s) of x on the curve y=−8+90x^3−3x^5 does the tangent line have the largest slope?

slope = m = dy/dx = 270 x^2 -15 x^4

where is m max?
dm/dx = 0 for max or min
dm/dx = 540 x - 60 x^3 = 0
x (540 - 60x^2) = 0
x = 0 or x = +/- 3

max or min at x = 0
d^2m/dx^2 = 540 - 180 x^2
that is + if x = 0, thus minium
if x = +3 that is - for x = +or - 3 so maximums there
which is bigger?
540(-3)-60(-3)^3 = 0
540(+3)-60(3)^3 = 0
so equal maximum slopes at + and - 3

Well, finding the value of x where the tangent line has the largest slope sounds like a mathematical quest for dominance! It's like finding the king of slopes, huh?

To determine the largest slope, we need to find the derivative of the curve. Let's do some calculus, shall we? Taking the derivative of y with respect to x, we get:

dy/dx = 270x^2 - 15x^4

Now, we need to find the critical points where the slope could be maximized. We set the derivative equal to zero and solve for x:

270x^2 - 15x^4 = 0

Dividing both sides by 15x^2, we get:

18 - x^2 = 0

Rearranging a bit further, we find:

x^2 = 18

Hmm, square roots! So, the possible values of x where the tangent line has the largest slope are x = ±√18.

Now, to pick the winner among these contenders, we can evaluate the second derivative. However, I'm afraid this is where my clownish math skills reach their limit. Nonetheless, give it a try and remember, even math can be a circus sometimes!

To find the value(s) of x on the given curve where the tangent line has the largest slope, we need to find the maximum value of the derivative of the curve.

First, let's find the derivative of the curve y = -8 + 90x^3 - 3x^5. Using the power rule, the derivative is given by:

dy/dx = 0 + 270x^2 - 15x^4

Now, we need to find the critical points of the derivative by setting dy/dx = 0 and solving for x:

0 + 270x^2 - 15x^4 = 0

Factoring out x^2, we get:

x^2(270 - 15x^2) = 0

Setting each factor equal to zero, we have two possible solutions:

x^2 = 0 --> x = 0

270 - 15x^2 = 0

Solving the second equation, we get:

270 - 15x^2 = 0
15x^2 = 270
x^2 = 18
x = ±√18

So, we have three possible values of x: x = 0, x = √18, and x = -√18.

Next, we need to determine which of these values gives the maximum slope. We can do this by evaluating the second derivative of the curve at each critical point.

The second derivative of the curve is given by:

d^2y/dx^2 = 540x - 60x^3

Evaluating the second derivative at x = 0, we get:

d^2y/dx^2 = 0

This indicates that x = 0 is a point of inflection rather than a maximum or minimum. So, we need to evaluate the second derivative at x = √18 and x = -√18:

d^2y/dx^2 = 540√18 - 60(√18)^3
= 540√18 - 60(√18)^3
≈ 108.8

The second derivative at x = -√18 will be the same with the opposite sign:

d^2y/dx^2 = -108.8

Since the second derivative is positive at x = √18 and negative at x = -√18, the tangent line will have the largest slope at x = √18.

Therefore, the value(s) of x on the given curve where the tangent line has the largest slope is x = √18.

To find the value(s) of x on the curve y = -8 + 90x^3 - 3x^5 where the tangent line has the largest slope, we need to calculate the derivative of the given function and determine where it is equal to zero.

Let's start by finding the derivative of the function y = -8 + 90x^3 - 3x^5. To do this, we differentiate each term with respect to x:

The derivative of the constant term -8 is 0 since the derivative of any constant is zero.

The derivative of 90x^3 is 270x^2. (Using the power rule, we multiply the exponent of x by the coefficient, which in this case is 3 * 90 = 270.)

The derivative of 3x^5 is 15x^4. (Again using the power rule, we multiply the exponent of x by the coefficient, which in this case is 5 * 3 = 15.)

Now that we have the derivative, which is dy/dx = 270x^2 + 15x^4, we need to find the critical points. These are the points where the derivative is equal to zero:

270x^2 + 15x^4 = 0

Factoring out x^2, we get:

x^2(270 + 15x^2) = 0

This equation will be equal to zero if either x^2 = 0 or (270 + 15x^2) = 0:

1. x^2 = 0
Solving for x, we find x = 0.

2. 270 + 15x^2 = 0
Subtracting 270 from both sides, we get 15x^2 = -270.
Dividing both sides by 15, we have x^2 = -18.
Since there is no real number solution for x^2 = -18, this case does not provide a critical point.

Therefore, the only critical point is x = 0.

The tangent line with the largest slope will occur at this critical point x = 0. This is because the derivative is zero at this point and changes sign around it, indicating a maximum or minimum point.

Thus, the value of x where the tangent line has the largest slope on the curve y = -8 + 90x^3 - 3x^5 is x = 0.