Question

1. consider the graph of the function ( f(x)=x^{3}-5 x^{2}-2 x + 24 ).

a) label the local minimum as a on the graph.
b) label the local maximum as b on the graph.
c) write down the interval where ( f^{prime}(x)>0 ).
d) draw the tangent to the curve at ( x = 0 ) on the graph.
e) write down the equation of the tangent at ( x = 0 ).

Explanation:

Step1: Find the derivative of \(f(x)\)

The function is \(f(x)=x^{3}-5x^{2}-2x + 24\). Using the power rule \((x^{n})^\prime=nx^{n - 1}\), we have \(f^\prime(x)=3x^{2}-10x-2\).

Step2: Analyze the sign of \(f^\prime(x)\)

We can find the roots of \(f^\prime(x)=3x^{2}-10x - 2\) using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c = 0\). Here \(a = 3\), \(b=-10\), \(c=-2\).
\(x=\frac{10\pm\sqrt{100+24}}{6}=\frac{10\pm\sqrt{124}}{6}=\frac{10\pm2\sqrt{31}}{6}=\frac{5\pm\sqrt{31}}{3}\approx\frac{5\pm5.57}{3}\). The roots are \(x_1=\frac{5 + \sqrt{31}}{3}\approx3.52\) and \(x_2=\frac{5-\sqrt{31}}{3}\approx - 0.19\).
The parabola \(y = f^\prime(x)=3x^{2}-10x - 2\) (since \(a = 3>0\)) opens upwards. The function \(y = f(x)\) is increasing when \(f^\prime(x)>0\). The solution of \(3x^{2}-10x - 2>0\) is \(x<\frac{5-\sqrt{31}}{3}\) or \(x>\frac{5+\sqrt{31}}{3}\). In interval notation, \((-\infty,\frac{5 - \sqrt{31}}{3})\cup(\frac{5+\sqrt{31}}{3},+\infty)\).

Step3: Find the equation of the tangent at \(x = 0\)

First, find \(f(0)\): \(f(0)=0^{3}-5\times0^{2}-2\times0 + 24=24\).
Then find \(f^\prime(0)\): \(f^\prime(x)=3x^{2}-10x-2\), so \(f^\prime(0)=- 2\).
Using the point - slope form of a line \(y - y_1=m(x - x_1)\) (where \((x_1,y_1)=(0,24)\) and \(m=-2\)), the equation of the tangent is \(y-24=-2(x - 0)\).

Answer:

c) \((-\infty,\frac{5-\sqrt{31}}{3})\cup(\frac{5+\sqrt{31}}{3},+\infty)\)
e) \(y=-2x + 24\)