Question

1. maximum mark: 8 a curve has equation ( y = h(x) ). given that - ( h(x) ) is a cubic expression whose coefficient of ( x^{3} ) is twice the coefficient of ( x ) - the curve passes through the point ( (-1,5) ) - the curve has a stationary point at ( (1,4) ) (a) find ( h(x) ). (b) determine whether the stationary point at ( (1,4) ) is a maximum or a minimum.

Explanation:

Step1: Set up the general form of the cubic function

Let \(h(x)=ax^{3}+bx^{2}+cx + d\). Then \(h^{\prime}(x)=3ax^{2}+2bx + c\).
Since the coefficient of \(x^{3}\) is twice the coefficient of \(x\), we have \(a = 2c\).

Step2: Use the point \((-1,5)\)

Substitute \(x=-1\) and \(y = 5\) into \(h(x)\):
\(h(-1)=-a + b - c + d=5\).

Step3: Use the stationary - point \((1,4)\)

Substitute \(x = 1\) and \(y = 4\) into \(h(x)\): \(h(1)=a + b + c + d=4\).
Substitute \(x = 1\) into \(h^{\prime}(x)\) (since at stationary - point \(h^{\prime}(x)=0\)): \(h^{\prime}(1)=3a + 2b + c=0\).

Step4: Solve the system of equations

We have the system of equations:
\(

$$\begin{cases}a = 2c\\-a + b - c + d=5\\a + b + c + d=4\\3a + 2b + c=0\end{cases}$$

\)
Subtract the second equation from the third equation: \((a + b + c + d)-(-a + b - c + d)=4 - 5\), which gives \(2a+2c=-1\).
Since \(a = 2c\), substitute \(a\) in \(2a + 2c=-1\): \(2(2c)+2c=-1\), \(4c+2c=-1\), \(6c=-1\), \(c=-\frac{1}{6}\), then \(a=-\frac{1}{3}\).
Substitute \(a\) and \(c\) into \(3a + 2b + c=0\): \(3(-\frac{1}{3})+2b-\frac{1}{6}=0\), \(-1+2b-\frac{1}{6}=0\), \(2b=\frac{7}{6}\), \(b=\frac{7}{12}\).
Substitute \(a\), \(b\), \(c\) into \(a + b + c + d=4\): \(-\frac{1}{3}+\frac{7}{12}-\frac{1}{6}+d=4\).
\(-\frac{4}{12}+\frac{7}{12}-\frac{2}{12}+d=4\), \(\frac{-4 + 7-2}{12}+d=4\), \(\frac{1}{12}+d=4\), \(d=\frac{47}{12}\).
So \(h(x)=-\frac{1}{3}x^{3}+\frac{7}{12}x^{2}-\frac{1}{6}x+\frac{47}{12}\).

Step5: Determine the nature of the stationary - point

Find the second - derivative \(h^{\prime\prime}(x)=6ax+2b\).
Substitute \(a =-\frac{1}{3}\) and \(b=\frac{7}{12}\) into \(h^{\prime\prime}(x)\): \(h^{\prime\prime}(x)=6(-\frac{1}{3})x+2(\frac{7}{12})=-2x+\frac{7}{6}\).
Evaluate \(h^{\prime\prime}(1)\): \(h^{\prime\prime}(1)=-2\times1+\frac{7}{6}=-\frac{12}{6}+\frac{7}{6}=-\frac{5}{6}<0\).

Answer:

(a) \(h(x)=-\frac{1}{3}x^{3}+\frac{7}{12}x^{2}-\frac{1}{6}x+\frac{47}{12}\)
(b) Since \(h^{\prime\prime}(1)=-\frac{5}{6}<0\), the stationary point at \((1,4)\) is a maximum.