Question

Use the First Derivative Test to find the location of all local extrema for the function given below.

f\left(x\right)\:=\:10\:+\:x^3-\:\frac{9x^2}{2}
Enter an exact
answer. If there is more than one local maximum or minimum, write each value of separated by a comma. If a
local maximum or local minimum does not occur on the function, enter in the appropriate box.

Answer 1

x = 0 is a local maximum and x = 3 is a local minimum.

Explanation:

To solve for the local extrema of the given function using the First Derivative Test, we'll first find the critical points by differentiating the function, then determine whether these points are local maxima or minima.

Given function:

`f(x)=10+x^3-(9x^2)/(2)`

`\hrulefill`

Differentiating the Function: The first derivative of f(x) is found by differentiating each term separately:

`\Longrightarrow (d)/(dx)\left[f(x)=10+x^3-(9x^2)/(2)\right]\\\\\\\\\Longrightarrow (d)/(dx)\left[f(x)\right]= (d)/(dx)\left[10\right]+ (d)/(dx)\left[x^3\right]- (d)/(dx)\left[(9x^2)/(2)\right]\\\\\\\\\Longrightarrow (d)/(dx)\left[f(x)\right]= (d)/(dx)\left[10\right]+ (d)/(dx)\left[x^3\right]- (9)/(2)(d)/(dx)\left[x^2\right]`

We can use the power rule to differentiate each term:

`\boxed{\begin{array}{ccc}\text{\underline{Power Rule for Differentiation:}}\\\\ (d)/(dx)[x^n]=nx^(n-1) \\ &\end{array}}`

`\Longrightarrow f'(x)= 0+3x^2-2\cdot (9)/(2)x\\\\\\\\\therefore f'(x)=3x^2-9x`

Finding Critical Points: To find critical points, we set the first derivative equal to zero:

`\Longrightarrow 0=3x^2-9x\\\\\\\\\Longrightarrow 0=x(3x-9)\\\\\\\\\therefore x=0 \text{ and } 3x-9=0`

Solving 3x - 9 = 0, we get:

`\Longrightarrow 3x-9=0\\\\\\\\\Longrightarrow 3x=9\\\\\\\\\therefore x =3`

Thus, we have two critical points: x = 0 and x = 3.

Applying the First Derivative Test: We'll examine the sign of f(x) around each critical point to determine the nature of each.

• For x = 0; Choose test points like -1 and 1.

When x = -1:

⇒ f'(-1) = 3(-1)² - 9(-1)

⇒ f'(-1) = 3 + 9

∴ f'(-1) = 12; f'(-1) > 0

When x = 1:

⇒ f'(1) = 3(1)² - 9(1)

⇒ f'(1) = 3 - 9

∴ f'(1) = -6; f'(1) < 0

Since the derivative changes from positive to negative, x = 0 is a local maximum.

• For x = 3; Choose test points like 2 and 4.

When x = 2:

⇒ f'(2) = 3(2)² - 9(2)

⇒ f'(2) = 3(4) - 18

⇒ f'(2) = 12 - 18

∴ f'(2) = -6; f'(2) < 0

When x = 4:

⇒ f'(4) = 3(4)² - 9(4)

⇒ f'(4) = 3(16) - 36

⇒ f'(4) = 48 - 36

∴ f'(4) = 12; f'(4) > 0

Since the derivative changes from negative to positive, x = 3 is a local minimum.