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.
\[
g(x)=x^{3}-9 x^{2}+15 x+1
\]
Answer
Finally, we can sketch the graph of the function using the information we have gathered. The function has a relative maximum at \(x = 1\), a relative minimum at \(x = 5\), and a point of inflection at \(x = 3\). The function is increasing on the intervals \((-\infty, 1)\) and \((5, \infty)\), and decreasing on the interval \((1, 5)\). The function is concave down on the interval \((-\infty, 3)\) and concave up on the interval \((3, \infty)\).
Step 1 ：First, we need to find the derivative of the function \(g(x) = x^{3} - 9x^{2} + 15x + 1\). The derivative of a function gives us the slope of the tangent line at any point, which can help us identify critical points and determine whether the function is increasing or decreasing.
Step 2 ：The derivative of \(g(x)\) is \(g'(x) = 3x^{2} - 18x + 15\).
Step 3 ：To find the critical points, we set the derivative equal to zero and solve for \(x\). So, we have \(3x^{2} - 18x + 15 = 0\).
Step 4 ：Factoring the equation gives us \(3(x - 1)(x - 5) = 0\). Setting each factor equal to zero gives us \(x = 1\) and \(x = 5\). So, the critical points are at \(x = 1\) and \(x = 5\).
Step 5 ：To classify each critical point as a relative maximum, a relative minimum, or neither, we can use the second derivative test. The second derivative of \(g(x)\) is \(g''(x) = 6x - 18\).
Step 6 ：Substituting \(x = 1\) into \(g''(x)\), we get \(g''(1) = -12\), which is less than zero. Therefore, \(x = 1\) is a relative maximum.
Step 7 ：Substituting \(x = 5\) into \(g''(x)\), we get \(g''(5) = 12\), which is greater than zero. Therefore, \(x = 5\) is a relative minimum.
Step 8 ：To identify intervals where the function is increasing or decreasing, we can use the first derivative. The function is increasing where the first derivative is positive and decreasing where the first derivative is negative.
Step 9 ：Substituting \(x = 0\) into \(g'(x)\), we get \(g'(0) = 15\), which is greater than zero. Therefore, the function is increasing on the interval \((-\infty, 1)\).
Step 10 ：Substituting \(x = 2\) into \(g'(x)\), we get \(g'(2) = -3\), which is less than zero. Therefore, the function is decreasing on the interval \((1, 5)\).
Step 11 ：Substituting \(x = 6\) into \(g'(x)\), we get \(g'(6) = 3\), which is greater than zero. Therefore, the function is increasing on the interval \((5, \infty)\).
Step 12 ：To find the points of inflection, we set the second derivative equal to zero and solve for \(x\). So, we have \(6x - 18 = 0\). Solving for \(x\) gives us \(x = 3\). So, the point of inflection is at \(x = 3\).
Step 13 ：To identify intervals where the function is concave up or concave down, we can use the second derivative. The function is concave up where the second derivative is positive and concave down where the second derivative is negative.
Step 14 ：Substituting \(x = 2\) into \(g''(x)\), we get \(g''(2) = -6\), which is less than zero. Therefore, the function is concave down on the interval \((-\infty, 3)\).
Step 15 ：Substituting \(x = 4\) into \(g''(x)\), we get \(g''(4) = 6\), which is greater than zero. Therefore, the function is concave up on the interval \((3, \infty)\).
Step 16 ：Finally, we can sketch the graph of the function using the information we have gathered. The function has a relative maximum at \(x = 1\), a relative minimum at \(x = 5\), and a point of inflection at \(x = 3\). The function is increasing on the intervals \((-\infty, 1)\) and \((5, \infty)\), and decreasing on the interval \((1, 5)\). The function is concave down on the interval \((-\infty, 3)\) and concave up on the interval \((3, \infty)\).
