Since you are MAXIMIZING the slope of the tangent line (which is represented by the first derivative), you need to look at the second derivative for the maximums of the first derivative.
y' = 180x² - 10x⁴
y" = 360x - 40x³
0 = 360x - 40x³
0 = 40x(9-x²) = 40x(3-x)(3+x)
Critical values at x=0,3,-3. Let's test the intervals created by these critical values:
( -∞, -3) :: test value = -4 :: y" > 0 → y' increasing
( -3, 0) :: test value = -2 :: y" < 0 → y' decreasing
(0, 3) :: test value = 1 :: y" > 0 → y' increasing
(3,∞) :: test value = 4 :: y" < 0 → y' decreasing
y' has a relative maximum when y' changes from increasing to decreasing, which is at x = -3, 3. The points are:
(-3, -1133) & (3, 1135), and that maximum slope is y'(±3) = 810.
dy/dx = 180x^2 -10x^4 = 0 ====> x = 0 and x = +3sqrt(2) and x = -3sqrt(2). Now for x = 0 the tangent line slope is 180(0) - 10(0) = 0. For x = +3sqrt(2) and -3sqrt(2) the tangent line slope is 180(18) - 10(18^2) = 2916, therefore the point waned is x = plus minus 3sqrt(2).
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slope is derivative y ' :
y ' = -10 x^4 + 180 x²
We search the maximum value, so we derive again :
y " = - 40 x³ + 360 x = 0
=> y " = 40 x (9 - x²) = 0
=> x = +- 3 or 0
slope for x = 0 is zero, and this is a LOCAL minimum (value = 0)
slope for x = +- 3 is 810 and this is the GLOBAL maximum.
corresponding y-value is :
for x = -3 : 1 - 60*27 + 2*243 = -1133
for x = 3 : 1 + 60 * 27 - 2*243 = 1135
So this occurs for the following points :
(-3, - 1133) and
(3, 1135)
y = 1 + 60x^3 − 2x^5
the slope of this curve is equal to dy/dx
=> dy/dx = d/dx(1) + d/dx(60x^3) − d/dx(2x^5)
=> dy/dx = 180x^2 − 10x^4
now local extrema of dy/dx are where d²y/dx² = 0
=> d²y/dx² = d/dx(180x^2) − d/dx(10x^4)
=> d²y/dx² = 360x − 40x^3
now 0 = 360x − 40x^3
=> 0 = 40x(9 - x^2)
=> 0 = 40x(3 + x)(3 - x)
so the extrema of dy/dx are at x = {-3, 0, 3}
extrema of dy/dx are local maximums when d³y/dx³ < 0 [ and local minimums at d³y/dx³ > 0]
=> d³y/dx³ = d/dx(360x) − d/dx(40x^3)
=> d³y/dx³ = 360 − 120x^2
now 360 − 120(-3)^2 = -720
360 − 120(0)^2 = 360
360 − 120(3)^2 = -720
so dy/dx is at maximum slope at x = {-3, 3}
y = 1 + 60(-3)^3 − 2(-3)^5 = -1133
y = 1 + 60(3)^3 − 2(3)^5 = 1135
so the two tangent points the resulting is the largest slopes are (-3,-1133) and (3,1135)
Since you are MAXIMIZING the slope of the tangent line (which is represented by the first derivative), you need to look at the second derivative for the maximums of the first derivative.
y' = 180x² - 10x⁴
y" = 360x - 40x³
0 = 360x - 40x³
0 = 40x(9-x²) = 40x(3-x)(3+x)
Critical values at x=0,3,-3. Let's test the intervals created by these critical values:
( -∞, -3) :: test value = -4 :: y" > 0 → y' increasing
( -3, 0) :: test value = -2 :: y" < 0 → y' decreasing
(0, 3) :: test value = 1 :: y" > 0 → y' increasing
(3,∞) :: test value = 4 :: y" < 0 → y' decreasing
y' has a relative maximum when y' changes from increasing to decreasing, which is at x = -3, 3. The points are:
(-3, -1133) & (3, 1135), and that maximum slope is y'(±3) = 810.
dy/dx = 180x^2 -10x^4 = 0 ====> x = 0 and x = +3sqrt(2) and x = -3sqrt(2). Now for x = 0 the tangent line slope is 180(0) - 10(0) = 0. For x = +3sqrt(2) and -3sqrt(2) the tangent line slope is 180(18) - 10(18^2) = 2916, therefore the point waned is x = plus minus 3sqrt(2).