Question

point of inflexion of the curve Y equal to x cube​

Answer 1

Answer:

Step-by-step explanation:

The first derivative of  y=x3  is  y′=3x2 . The second derivative is  y′′=6x .

If  6x=0 ,  x=0 . So the possible point of inflection is  (0,0) .

Then, if  (0,0)  is an inflection point, the function will be concave up to one side and concave down to the other. We check the concavity on either side by plugging some value less than 0 and some value greater than 0 into the second derivative. If the result is less than 0, it is concave down; if greater, concave up. I will test the values -1 and 1.

y′′(−1)=6(−1)=−6<0 . To the left of (0,0), the function is concave down.

y′′(1)=6(1)=6>0 . To the right of (0,0), the function is concave down.

The point of inflection of the graph of  y=x3  is  (0,0) .

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