To find the relative extrema of the function Take the first derivative of the function:<\/p>\n\n\n\n f'(x) = d\/dx [x\u00b3 \u2212 6x\u00b2 + 6] = 3x\u00b2 \u2212 12x<\/strong><\/p>\n\n\n\n Set the first derivative equal to zero to find critical points:<\/p>\n\n\n\n 3x\u00b2 \u2212 12x = 0<\/strong><\/p>\n\n\n\n Factor:<\/p>\n\n\n\n 3x(x \u2212 4) = 0<\/strong><\/p>\n\n\n\n So the critical points are:<\/p>\n\n\n\n x = 0<\/strong> and x = 4<\/strong><\/p>\n\n\n\n Find the second derivative:<\/p>\n\n\n\n f”(x) = d\/dx [3x\u00b2 \u2212 12x] = 6x \u2212 12<\/strong><\/p>\n\n\n\n Evaluate the second derivative at each critical point.<\/p>\n\n\n\n To identify the relative extrema of a function, we look for critical points where the first derivative is zero or undefined. For the function f(x) = x\u00b3 \u2212 6x\u00b2 + 6, the first derivative is f'(x) = 3x\u00b2 \u2212 12x. Setting this equal to zero gives us the critical points: x = 0 and x = 4.<\/p>\n\n\n\n Next, we use the second derivative test to classify these critical points. The second derivative, f”(x) = 6x \u2212 12, helps us determine concavity. If the second derivative is positive at a critical point, the graph is concave up, and the point is a relative minimum. If the second derivative is negative, the graph is concave down, and the point is a relative maximum.<\/p>\n\n\n\n Evaluating the second derivative at x = 0, we find f”(0) = \u221212, which is negative. This indicates that the function is concave down at x = 0, so this point is a relative maximum. At x = 4, f”(4) = 12, which is positive, showing that the function is concave up, and the point is a relative minimum.<\/p>\n\n\n\n Finally, we substitute these x-values into the original function to find their corresponding y-values. At x = 0, f(0) = 6, and at x = 4, f(4) = \u221226. So the relative maximum is at (0, 6), and the relative minimum is at (4, \u221226).<\/p>\n\n\n\n This analysis shows how calculus tools like the first and second derivatives help identify and classify relative extrema in polynomial functions, giving valuable information about the graph\u2019s shape and turning points.<\/p>\n\n\n\n Find all relative extrema of the function. Use the Second Derivative Test where applicable. (If an answer does not exist, enter DNE:) f(x) = x3 6×2 + 6 relative maximum (x, Y) = ( relative minimum (x,Y) = The Correct Answer and Explanation is: To find the relative extrema of the functionf(x) = x\u00b3 \u2212 […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-34985","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/34985","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=34985"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/34985\/revisions"}],"predecessor-version":[{"id":34995,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/34985\/revisions\/34995"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=34985"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=34985"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=34985"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
f(x) = x\u00b3 \u2212 6x\u00b2 + 6<\/strong>,
we follow these steps:<\/p>\n\n\n\n
\n\n\n\nStep 1: Find the critical points<\/strong><\/h3>\n\n\n\n
\n\n\n\nStep 2: Apply the Second Derivative Test<\/strong><\/h3>\n\n\n\n
\n
f”(0) = 6(0) \u2212 12 = \u221212<\/strong>
Since this is less than zero, the function is concave down at x = 0.
Therefore, there is a relative maximum<\/strong> at x = 0<\/strong>.<\/li>\n\n\n\n
f”(4) = 6(4) \u2212 12 = 24 \u2212 12 = 12<\/strong>
Since this is greater than zero, the function is concave up at x = 4.
Therefore, there is a relative minimum<\/strong> at x = 4<\/strong>.<\/li>\n<\/ul>\n\n\n\n
\n\n\n\nStep 3: Find the function values at those points<\/strong><\/h3>\n\n\n\n
\n
\n\n\n\nFinal Answer:<\/strong><\/h3>\n\n\n\n
\n
\n\n\n\nExplanation <\/strong><\/h3>\n\n\n\n
![\"\"](\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-613.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"