Question

Find the local maximum and minimum values of f using both the first and second derivative tests. f(x) = 6 9x2 − 6x3 local maximum value local minimum value

Answer 1

The local minimum value of the function f(x) = 6 + 9x² - 6x³ is 6 and the local maximum value of the function f(x) = 6 + 9x² - 6x³ is 9.

For given question,

We have been given a function f(x) = 6 + 9x² - 6x³

We need to find the local maximum and local minimum of the function  f(x)

First we find the first derivative of the function.

⇒ f'(x) = 0 + 18x - 18x²

⇒ f'(x) = - 18x² + 18x

Putting the first derivative of the function equal to zero, we get

⇒ f'(x) = 0

⇒ - 18x² + 18x = 0

⇒ 18(-x² + x) = 0

⇒ x (-x + 1) = 0

⇒ x = 0    or    -x + 1 = 0

⇒ x = 0     or    x = 1

Now we find the second derivative of the function.

⇒ f"(x) = - 36x + 18

At x = 0 the value of second derivative of function f(x),

⇒ f"(0) = - 36(0) + 18

⇒ f"(0) = 0 + 18

⇒ f"(0) = 18

Here, at x=0, f"(x) > 0

This means, the function f(x) has the local minimum value at x = 0,  which is given by

⇒ f(0) = 6 + 9(0)² - 6(0)³

⇒ f(0) = 6 + 0 - 0

⇒ f(0) = 6

At x = 1 the value of second derivative of function f(x),

⇒ f"(1) = - 36(1) + 18

⇒ f"(1) = - 18

Here, at x = 1, f"(x) < 0

This means, the function f(x) has the local maximum value at x = 1,  which is given by

⇒ f(1) = 6 + 9(1)² - 6(1)³

⇒ f(1) = 6 + 9 - 6

⇒ f(1) = 9

So, the function f(x) = 6 + 9x² - 6x³ has local minimum at x = 0 and local maximum at x = 1.

Therefore, the local minimum value of the function f(x) = 6 + 9x² - 6x³ is 6 and the local maximum value of the function f(x) = 6 + 9x² - 6x³ is 9.

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