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Respond to the following questions for the function y = x³(2-x)².

(a) (dy/dx) at (x=1) = ?
(b) The point on the graph for x=1 is?
(c) Equation of the tangent at

User Brillian
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1 Answer

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Final answer:

The derivative and point on the graph are calculated as 5, and the equation of tangent is determined as y - 1 = 5(x - 1).

Step-by-step explanation:

(a) To find (dy/dx) at (x=1), we need to differentiate the function y = x³(2-x)² with respect to x and then substitute x=1 into the derivative. Let's find the derivative step by step:

Using the product rule, differentiate x³ and (2-x)² separately. The derivative of x³ is 3x² and the derivative of (2-x)² is -2(2-x).

Now, apply the chain rule to the term (2-x), which gives us (-1)*(-1) = 1. Therefore, the derivative of y = x³(2-x)² is 3x²(2-x) + x³(-2)(-1) = 3x²(2-x) + 2x³.

Now, substitute x=1 into the derivative: (dy/dx) at (x=1) = 3(1)²(2-1) + 2(1)³ = 3(1)(1) + 2(1) = 3 + 2 = 5.

(b) To find the point on the graph for x=1, substitute x=1 into the original function: y = (1)³(2-1)² = 1(1)(1) = 1.

(c) To find the equation of the tangent at x=1, we need to use the point-slope form of the equation for a line. We have the point (1,1) and the slope (dy/dx) at (x=1) = 5 from part (a). So, the equation of the tangent is y - 1 = 5(x - 1).

User Urbz
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