Question

The graph of the function y = x^3 + ax^2 + b has a local minimum point at (4, -11). Find the coordinates of the local maximum. Answer: (0, 21)

Answer 1

Answer:

(0, 21)

Step-by-step explanation:

Stationary points occur when the gradient of a graph is zero.

Therefore, to find the x-coordinate(s) of the stationary points of a function, differentiate the function, set it to zero and solve for x.

Differentiate the given function:

$\begin{aligned}y & = x^3+ax^2+b\\\implies \dfrac{\text{d}y}{\text{d}x}&=3 \cdot x^{3-1}+2 \cdot ax^{2-1}+0\\\implies \dfrac{\text{d}y}{\text{d}x}& = 3x^2+2ax\end{aligned}$

Set the differentiated function to zero:

$\begin{aligned} \dfrac{\text{d}y}{\text{d}x}& =0\\ \implies 3x^2+2ax & = 0 \\ \implies x(3x+2a)&=0\end{aligned}$

Therefore the stationary points occur when:

$\implies x=0$

$\implies 3x+2a=0 \implies x=-\dfrac{2}{3}a$

There is a stationary point at (4, -11), therefore substitute x = 4 into the expression for x and solve for a:

$\begin{aligned}x & = 4\\\implies -\dfrac{2}{3}a & = 4\\a & = -\dfrac{3}{2}(4)\\a & = -6\end{aligned}$

Substitute the found value of a and the point (4, -11) into the function and solve for b:

$\begin{aligned}y & = x^3 + ax^2 + b\\\implies -11 & = (4)^3+(-6)(4)^2+b\\ -11 & = 64-96+b\\b & = -11-64+96\\ b & = 21\end{aligned}$

Therefore, the function is:

$\boxed{y=x^3-6x^2+21}$

The other stationary point is when x = 0.  Therefore, to find the coordinates of this point, substitute x = 0 into the function:

$\begin{aligned}y & =x^3-6x^2+21\\x=0\implies y & =(0)^3-6(0)^2+21\\ y & =21 \\end{aligned}$

Therefore, the coordinates of the other stationary point are (0, 21).

To determine if a stationary point is minimum or maximum, differentiate the function again:

$\begin{aligned}y & =x^3-6x^2+21\\\implies \dfrac{\text{d}y}{\text{d}x} & = 3x^2-12x\\\implies \dfrac{\text{d}^2y}{\text{d}x^2} & = 6x-12\\\end{aligned}$

Substitute the x-coordinate of the stationary point into the second derivative:

$\begin{aligned} \dfrac{\text{d}^2y}{\text{d}x^2} & = 6x-12\\x=0 \implies \dfrac{\text{d}^2y}{\text{d}x^2} & = 6(0)-12= -12 \\\end{aligned}$

$\textsf{As \dfrac{\text{d}^2y}{\text{d}x^2} < 0 , stationary point (0,21) is a maximum.}$