Question

f(x) = x^3 -4x^2 -11x + 30
Determine the stationary point of the graph of y = f’(x)

Answer 1

Answer:

$\left(\dfrac{4}{3},-\dfrac{49}{3}\right)$

Step-by-step explanation:

A stationary point on a graph is a point where the derivative of the function is zero, indicating a possible maximum, minimum, or point of inflection.

The x-coordinates of the stationary points of the graph of f(x) are the x-values when the derivative of the function equals zero. Therefore, the x-coordinate of the stationary point of the graph of y = f'(x) is the x-value when f''(x) = 0.

Differentiate f(x) to find f'(x):

$f'(x)=3 \cdot x^{3-1} - 2 \cdot 4x^{2-1}-1 \cdot 11x^{1-1}+0\\\\f'(x)=3x^2-8x-11$

Differentiate f'(x) to find f''(x):

$f''(x)=2 \cdot 3x^{2-1}-1 \cdot 8x^{1-1}-0\\\\f''(x)=6x-8$

Set f''(x) equal to zero and solve for x:

$f''(x)=0\\\\\\6x-8=0\\\\\\6x=8\\\\\\x=\dfrac{4}{3}$

Therefore, the x-coordinate of the stationary point of the graph of y = f'(x) is x = 4/3.

To find the corresponding y-coordinate, substitute x = 4/3 into f'(x):

$f'\left(\dfrac{4}{3}\right)=3\left(\dfrac{4}{3}\right)^2-8\left(\dfrac{4}{3}\right)-11\\\\\\\\f'\left(\dfrac{4}{3}\right)=\dfrac{16}{3}-\dfrac{32}{3}-\dfrac{33}{3}\\\\\\\\f'\left(\dfrac{4}{3}\right)=-\dfrac{49}{3}$

Therefore, the stationary point of the graph of y = f’(x) is:

$\left(\dfrac{4}{3},-\dfrac{49}{3}\right)$