To determine whether the derivative of f(x)=\u2223x+8\u2223f(x) = |x + 8| exists at x=\u22128x = -8, we must evaluate the left-hand and right-hand limits using the alternative form of the derivative:<\/p>\n\n\n\n
f\u2032(c)=lim\u2061x\u2192cf(x)\u2212f(c)x\u2212cf'(c) = \\lim_{{x \\to c}} \\frac{{f(x) – f(c)}}{{x – c}}<\/p>\n\n\n\n
Let c=\u22128c = -8. Then f(\u22128)=\u2223\u22128+8\u2223=\u22230\u2223=0f(-8) = | -8 + 8 | = |0| = 0.<\/p>\n\n\n\n
Left-hand derivative (as x\u2192\u22128\u2212x \\to -8^-):<\/h3>\n\n\n\n
When x<\u22128x < -8, x+8<0x + 8 < 0, so f(x)=\u2212(x+8)f(x) = -(x + 8). Thus,<\/p>\n\n\n\n
lim\u2061x\u2192\u22128\u2212f(x)\u2212f(\u22128)x+8=lim\u2061x\u2192\u22128\u2212\u2212(x+8)\u22120x+8=lim\u2061x\u2192\u22128\u2212\u2212(x+8)x+8=\u22121\\lim_{{x \\to -8^-}} \\frac{{f(x) – f(-8)}}{{x + 8}} = \\lim_{{x \\to -8^-}} \\frac{{-(x + 8) – 0}}{{x + 8}} = \\lim_{{x \\to -8^-}} \\frac{{-(x + 8)}}{{x + 8}} = -1<\/p>\n\n\n\n
Right-hand derivative (as x\u2192\u22128+x \\to -8^+):<\/h3>\n\n\n\n
When x>\u22128x > -8, x+8>0x + 8 > 0, so f(x)=x+8f(x) = x + 8. Thus,<\/p>\n\n\n\n
lim\u2061x\u2192\u22128+f(x)\u2212f(\u22128)x+8=lim\u2061x\u2192\u22128+x+8\u22120x+8=lim\u2061x\u2192\u22128+x+8x+8=1\\lim_{{x \\to -8^+}} \\frac{{f(x) – f(-8)}}{{x + 8}} = \\lim_{{x \\to -8^+}} \\frac{{x + 8 – 0}}{{x + 8}} = \\lim_{{x \\to -8^+}} \\frac{{x + 8}}{{x + 8}} = 1<\/p>\n\n\n\n
Conclusion:<\/h3>\n\n\n\n
The left-hand derivative is -1 and the right-hand derivative is 1. Since these two one-sided limits are not equal, the derivative at x=\u22128x = -8 does not<\/strong> exist.<\/p>\n\n\n\n This result is consistent with the nature of absolute value functions. The graph of f(x)=\u2223x+8\u2223f(x) = |x + 8| has a sharp corner at x=\u22128x = -8, which means the slope changes abruptly. A function is not differentiable at a point where it has a corner or cusp because the limit defining the derivative does not approach the same value from both sides. Therefore, the derivative of f(x)f(x) at x=\u22128x = -8 does not exist.<\/p>\n\n\n\n Alternative form of derivative Consider the derivative form above. Use the derivatives from the left and the right to find the derivative at x = -8 if it exists. (15 points) f(x) = |x + 8 The Correct Answer and Explanation is: To determine whether the derivative of f(x)=\u2223x+8\u2223f(x) = |x + 8| exists at […]<\/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-41074","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/41074","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=41074"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/41074\/revisions"}],"predecessor-version":[{"id":41076,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/41074\/revisions\/41076"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=41074"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=41074"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=41074"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}![\"\"](\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1317.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"