Question

for the following function, a) give the coordinates of any critical points and classify each point as a relative maximum, a relative minimum, or neither; b) identify intervals where the function is increasing or decreasing; c) give the coordinates of any points of inflection; d) identify intervals where the function is concave up or concave down, and e) sketch the graph. g(x)=x³ - 12x² + 45x + 2

Explanation:

Step1: Find the first derivative

The first derivative of \(g(x)=x^{3}-12x^{2}+45x + 2\) is \(g^{\prime}(x)=3x^{2}-24x + 45\). Factor it: \(g^{\prime}(x)=3(x^{2}-8x + 15)=3(x - 3)(x - 5)\).
Set \(g^{\prime}(x)=0\), then \(3(x - 3)(x - 5)=0\). Solving for \(x\), we get \(x = 3\) and \(x = 5\).
When \(x = 3\), \(g(3)=3^{3}-12\times3^{2}+45\times3+2=27-108 + 135+2=56\).
When \(x = 5\), \(g(5)=5^{3}-12\times5^{2}+45\times5+2=125-300+225 + 2=52\).
To classify the critical points, use the second - derivative test.
The second derivative \(g^{\prime\prime}(x)=6x-24\).
When \(x = 3\), \(g^{\prime\prime}(3)=6\times3-24=-6<0\), so \((3,56)\) is a relative maximum.
When \(x = 5\), \(g^{\prime\prime}(5)=6\times5-24 = 6>0\), so \((5,52)\) is a relative minimum.

Step2: Determine intervals of increase and decrease

We consider the intervals \((-\infty,3)\), \((3,5)\) and \((5,\infty)\).
Take a test point in \((-\infty,3)\), say \(x = 0\). Then \(g^{\prime}(0)=3\times(0 - 3)(0 - 5)=45>0\), so \(g(x)\) is increasing on \((-\infty,3)\).
Take a test point in \((3,5)\), say \(x = 4\). Then \(g^{\prime}(4)=3\times(4 - 3)(4 - 5)=-3<0\), so \(g(x)\) is decreasing on \((3,5)\).
Take a test point in \((5,\infty)\), say \(x = 6\). Then \(g^{\prime}(6)=3\times(6 - 3)(6 - 5)=9>0\), so \(g(x)\) is increasing on \((5,\infty)\).

Step3: Find the second derivative and points of inflection

Set \(g^{\prime\prime}(x)=0\), \(6x-24 = 0\), solving for \(x\) gives \(x = 4\).
When \(x = 4\), \(g(4)=4^{3}-12\times4^{2}+45\times4+2=64-192+180+2=54\). So the point of inflection is \((4,54)\).

Step4: Determine intervals of concavity

Consider the intervals \((-\infty,4)\) and \((4,\infty)\).
Take a test point in \((-\infty,4)\), say \(x = 3\). Then \(g^{\prime\prime}(3)=6\times3-24=-6<0\), so \(g(x)\) is concave down on \((-\infty,4)\).
Take a test point in \((4,\infty)\), say \(x = 5\). Then \(g^{\prime\prime}(5)=6\times5-24 = 6>0\), so \(g(x)\) is concave up on \((4,\infty)\).

Answer:

a) Critical points: \((3,56)\) (relative maximum), \((5,52)\) (relative minimum).
b) Increasing intervals: \((-\infty,3)\cup(5,\infty)\); Decreasing interval: \((3,5)\).
c) Point of inflection: \((4,54)\).
d) Concave down interval: \((-\infty,4)\); Concave up interval: \((4,\infty)\).
e) To sketch the graph: Plot the critical points \((3,56)\) (peak), \((5,52)\) (valley), the point of inflection \((4,54)\). Use the information about increasing/decreasing (rising on \((-\infty,3)\), falling on \((3,5)\), rising on \((5,\infty)\)) and concavity (concave down on \((-\infty,4)\), concave up on \((4,\infty)\)) to draw a smooth curve for \(y=x^{3}-12x^{2}+45x + 2\).