Final answer:
There are no values of x on the curve y=3+40x^3-3x^5 where the tangent line has the largest slope.
Step-by-step explanation:
To find the value(s) of x on the curve y=3+40x^3-3x^5 where the tangent line has the largest slope, we need to find the maximum slope of the tangent line. The slope of the tangent line to a curve is given by the derivative of the curve at that point. So, we need to find the derivative of the curve and set it equal to zero to find the critical points.
Let's differentiate the curve with respect to x:
dy/dx = 120x^2 - 15x^4
Setting the derivative equal to zero and solving for x gives us the critical points:
120x^2 - 15x^4 = 0
Factoring out 15x^2, we get:
15x^2(8 - x^2) = 0
So, either 15x^2 = 0 or 8 - x^2 = 0.
Solving for x gives us two critical points: x = 0 and x = ±√8.
To determine which critical point gives us the largest slope, we need to evaluate the second derivative at these critical points. The second derivative is given by:
d²y/dx² = 240x - 60x^3
Substituting the critical points:
For x = 0, d²y/dx² = 240(0) - 60(0)^3 = 0. For x = ±√8, d²y/dx² = 240(±√8) - 60(±√8)^3 = ±480√8 - 480√8 = 0.
Both critical points give us a second derivative of zero, which means they are inflection points and not points of maximum slope. Therefore, there are no values of x on the curve y=3+40x^3-3x^5 where the tangent line has the largest slope.