To find the value of the constant kkk that makes the function continuous at x=4x = 4x=4, we need to ensure that: lim\u2061x\u21924\u2212g(x)=g(4)=lim\u2061x\u21924+g(x)\\lim_{x \\to 4^-} g(x) = g(4) = \\lim_{x \\to 4^+} g(x)x\u21924\u2212lim\u200bg(x)=g(4)=x\u21924+lim\u200bg(x)<\/p>\n\n\n\n
The piecewise function is defined as:<\/p>\n\n\n\n
- \n
- g(x)=3×2\u22128x\u221216g(x) = 3x^2 – 8x – 16g(x)=3×2\u22128x\u221216 if x<4x < 4x<4<\/li>\n\n\n\n
- g(x)=kx\u22124g(x) = kx – 4g(x)=kx\u22124 if x=4x = 4x=4<\/li>\n\n\n\n
- g(x)=0g(x) = 0g(x)=0 if x>4x > 4x>4<\/li>\n<\/ul>\n\n\n\n
Step 1: Evaluate the limit as x\u21924\u2212x \\to 4^-x\u21924\u2212<\/h3>\n\n\n\n
Use the expression for x<4x < 4x<4: lim\u2061x\u21924\u2212g(x)=3(4)2\u22128(4)\u221216=48\u221232\u221216=0\\lim_{x \\to 4^-} g(x) = 3(4)^2 – 8(4) – 16 = 48 – 32 – 16 = 0x\u21924\u2212lim\u200bg(x)=3(4)2\u22128(4)\u221216=48\u221232\u221216=0<\/p>\n\n\n\n
Step 2: Evaluate the limit as x\u21924+x \\to 4^+x\u21924+<\/h3>\n\n\n\n
Use the expression for x>4x > 4x>4, which is a constant: lim\u2061x\u21924+g(x)=0\\lim_{x \\to 4^+} g(x) = 0x\u21924+lim\u200bg(x)=0<\/p>\n\n\n\n
Step 3: Since both one-sided limits equal 0:<\/h3>\n\n\n\n
lim\u2061x\u21924g(x)=0\\lim_{x \\to 4} g(x) = 0x\u21924lim\u200bg(x)=0<\/p>\n\n\n\n
To make the function continuous, the actual value at x=4x = 4x=4, that is g(4)g(4)g(4), must also equal 0. But from the definition: g(4)=k(4)\u22124=4k\u22124g(4) = k(4) – 4 = 4k – 4g(4)=k(4)\u22124=4k\u22124<\/p>\n\n\n\n
Set this equal to the limit value: 4k\u22124=04k – 4 = 04k\u22124=0<\/p>\n\n\n\n
Solve the equation:<\/h3>\n\n\n\n
4k=4k=14k = 4 \\\\ k = 14k=4k=1<\/p>\n\n\n\n
\n\n\n\nFinal Answer:<\/h3>\n\n\n\n
k=1k = 1k=1<\/strong><\/p>\n\n\n\n
\n\n\n\nExplanation <\/h3>\n\n\n\n
Continuity of a function at a specific point requires that the left-hand limit, the right-hand limit, and the function value at that point are all equal. In this problem, the function g(x)g(x)g(x) is piecewise-defined with different rules for when xxx is less than, equal to, or greater than 4.<\/p>\n\n\n\n
For the function to be continuous at x=4x = 4x=4, we must ensure:<\/p>\n\n\n\n
- \n
- lim\u2061x\u21924\u2212g(x)=lim\u2061x\u21924+g(x)\\lim_{x \\to 4^-} g(x) = \\lim_{x \\to 4^+} g(x)limx\u21924\u2212\u200bg(x)=limx\u21924+\u200bg(x)<\/li>\n\n\n\n
- g(4)g(4)g(4) is equal to that common limit value<\/li>\n<\/ul>\n\n\n\n
From the left, g(x)=3×2\u22128x\u221216g(x) = 3x^2 – 8x – 16g(x)=3×2\u22128x\u221216. Plugging in x=4x = 4x=4, we get: 3(4)2\u22128(4)\u221216=48\u221232\u221216=03(4)^2 – 8(4) – 16 = 48 – 32 – 16 = 03(4)2\u22128(4)\u221216=48\u221232\u221216=0<\/p>\n\n\n\n
From the right, g(x)=0g(x) = 0g(x)=0 for x>4x > 4x>4, so: lim\u2061x\u21924+g(x)=0\\lim_{x \\to 4^+} g(x) = 0x\u21924+lim\u200bg(x)=0<\/p>\n\n\n\n
Both limits equal 0, which means the overall limit exists and equals 0. Now, to make the function continuous, the function value at x=4x = 4x=4 must also be 0.<\/p>\n\n\n\n
From the definition, g(4)=k(4)\u22124=4k\u22124g(4) = k(4) – 4 = 4k – 4g(4)=k(4)\u22124=4k\u22124. Set this equal to the limit: 4k\u22124=0\u21d24k=4\u21d2k=14k – 4 = 0 \\Rightarrow 4k = 4 \\Rightarrow k = 14k\u22124=0\u21d24k=4\u21d2k=1<\/p>\n\n\n\n
Therefore, k=1k = 1k=1 ensures the function is continuous at x=4x = 4x=4.<\/p>\n\n\n\n
![\"\"](\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-129.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"Find the value of the constant k that makes the function continuous. 3x^2 – 8x – 16 if x < 4 kx – 4 if x = 4 g(x) Write an equation that can be solved to find k: 3x^2 – 8x – 16 = k(4) – 4 if x > 4 0 A. 842_84) […]<\/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-30208","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/30208","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=30208"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/30208\/revisions"}],"predecessor-version":[{"id":30214,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/30208\/revisions\/30214"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=30208"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=30208"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=30208"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}