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Flauer [41]
3 years ago
5

Sketch the curve y=x(x-2)^3 showing any stationary points and points of inflection

Mathematics
2 answers:
ycow [4]3 years ago
6 0

Answer:

hshsnwjsbsbskakdbdbsn

elixir [45]3 years ago
3 0

Answer:

Step-by-step explanation:

hello :

let   :   y=f(x)

f(x)==x(x-2)^3   so : f'(x) = 1.(x-2)^3+3x(x-2)²

f'(x) = (x-2)²(x-2+3x)

f'(x) = (x-2)²(4x-2)

f'(x) =0    : x=2  or x=1/2

the minumum is (1/2 ,f(2))

the inflection points use f"(x) = 0.......continu

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\text {x - coordinate = }  \bigg( 5\dfrac{1}{2}  +  3\dfrac{3}{4} \bigg) \div 2

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\text {x - coordinate = }\dfrac{37}{4}   \times  \dfrac{1}{2}

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