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Hint: To solve this question we will differentiate the given function two times. It is known as the second derivative test. If the value of the function comes out to be negative, then it is maximum for that particular value of x and if the value comes out to be positive, then it is minimum for that value of x.

Complete Step-by-step Answer:

Now, the given function is ${\text{y = }}{{\text{x}}^3}{\text{ - 3x + 2 }}$. We will use the second derivative test to find the local maxima for the given function. Now, there are two types of maxima and minima: Global and local. If the function has the largest value for a particular value of x, then the function is a global maxima for that value of x. If for another value of x, the function has value less than the global maxima, then it is local maxima for that x.

Now, differentiating both sides with respect to x, we get

$\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = 3}}{{\text{x}}^2}{\text{ - 3}}$ … (1)

Now, to find the values of x, we put $\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = 0}}$.

Therefore, $3{{\text{x}}^2}{\text{ - 3 = 0}}$

In the above equation, we will use the property ${{\text{x}}^2}{\text{ - }}{{\text{y}}^2}{\text{ = (x - y)(x + y)}}$, we get

${\text{3(x - 1)(x + 1) = 0}}$

Therefore, x = 1, -1

Now, differentiating equation (1) both sides with respect to x, we get

$\dfrac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}}{\text{ = 6x}}$

Now, putting x = 1 in the above equation, we get

$\dfrac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}}{\text{ = 6}}$

As, the double derivative is positive, so function is minimum at x = 1.

Now, putting x = -1, we get $\dfrac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}}{\text{ = - 6}}$

Now, the double derivative is negative, so the function has a maxima at x = -1.

So, option (D) is correct, i.e. function has a local maxima at x = -1.

Note: When we come up with such types of questions, we have to find the double derivative of the given function to solve the given problem. To find the value of x, we will put the first derivative equal to 0 and find the values of x from the equation formed. Then we will put all the values of x in the second derivative to check whether the double derivative is positive or negative. As soon as it comes to be negative, the function has a maxima on that value of x.

Complete Step-by-step Answer:

Now, the given function is ${\text{y = }}{{\text{x}}^3}{\text{ - 3x + 2 }}$. We will use the second derivative test to find the local maxima for the given function. Now, there are two types of maxima and minima: Global and local. If the function has the largest value for a particular value of x, then the function is a global maxima for that value of x. If for another value of x, the function has value less than the global maxima, then it is local maxima for that x.

Now, differentiating both sides with respect to x, we get

$\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = 3}}{{\text{x}}^2}{\text{ - 3}}$ … (1)

Now, to find the values of x, we put $\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = 0}}$.

Therefore, $3{{\text{x}}^2}{\text{ - 3 = 0}}$

In the above equation, we will use the property ${{\text{x}}^2}{\text{ - }}{{\text{y}}^2}{\text{ = (x - y)(x + y)}}$, we get

${\text{3(x - 1)(x + 1) = 0}}$

Therefore, x = 1, -1

Now, differentiating equation (1) both sides with respect to x, we get

$\dfrac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}}{\text{ = 6x}}$

Now, putting x = 1 in the above equation, we get

$\dfrac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}}{\text{ = 6}}$

As, the double derivative is positive, so function is minimum at x = 1.

Now, putting x = -1, we get $\dfrac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}}{\text{ = - 6}}$

Now, the double derivative is negative, so the function has a maxima at x = -1.

So, option (D) is correct, i.e. function has a local maxima at x = -1.

Note: When we come up with such types of questions, we have to find the double derivative of the given function to solve the given problem. To find the value of x, we will put the first derivative equal to 0 and find the values of x from the equation formed. Then we will put all the values of x in the second derivative to check whether the double derivative is positive or negative. As soon as it comes to be negative, the function has a maxima on that value of x.

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