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How do you find out the x-coordinates of the stationary points on the curve y equals (3x plus 1)caret5(3-x)caret3?
Wiki User
∙ 6y ago
y = (3x + 1)⁵(3 - x)³
The stationary points are where dy/dx = 0
Note the use of:
If u and v are functions of x:
d/dx (uv) = v du/dx + u dv/dx
d/dx uⁿ = n uⁿ⁻¹ du/dx
dy/dx = (3 - x)³ d/dx (3x + 1)⁵ + (3x + 1)⁵ d/dx (3 - x)³
= (3 - x)³ 5 (3x + 1)⁴ d/dx (3x + 1) + (3x + 1)⁵ 3 (3 - x)² d/dx (3 - x)
= 5 (3 - x)³ (3x + 1)⁴ 3 + 3 (3x + 1)⁵ (3 - x)² (-1)
= 3 (3x + 1)⁴ (3 - x)² (5 (3 - x) - (3x + 1))
= 3 (3x + 1)⁴ (3 - x)² (15 - 5x - 3x - 1)
= 3 (3x + 1)⁴ (3 - x)² (14 - 8x)
= 6 (3x + 1)⁴ (3 - x)² (7 - 4x)
→ 3x + 1 = 0 → x = -1/3
or 3 - x = 0 → x = 3
or 7 - 4x = 0 → x = 7/4
Wiki User
∙ 6y ago
Wiki User
∙ 6y agoy = (3x + 1)^5*(3 - x)^3dy/dx = 5* (3x + 1)^4*3*(3 - x)^3 - 3*(3x + 1)^5*(3 - x)^2
= 3*(3x + 1)^4*(3 - x)^2*[5*(3-x) - (3x + 1)]
= 6*(3x + 1)^4*(3 - x)^2*(7 - 4x)
Then the stationary points are reached when dy/dx = 0.
So, solving dy/dx = 0 gives
x = -1/3
x = 7/4
x = 3.
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