{"id":26437,"date":"2025-06-19T14:39:39","date_gmt":"2025-06-19T14:39:39","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=26437"},"modified":"2025-06-19T14:39:43","modified_gmt":"2025-06-19T14:39:43","slug":"use-the-alternate-form-of-the-derivative-to-find-the-derivative-at-x-c-if-it-exists","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/use-the-alternate-form-of-the-derivative-to-find-the-derivative-at-x-c-if-it-exists\/","title":{"rendered":"Use the alternate form of the derivative to find the derivative at x = c, if it exists."},"content":{"rendered":"\n
 f(x) = x^3 + 2x^2 + 1, c = -2<\/pre>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
We are asked to find the derivative of the function
f(x) = x\u00b3 + 2x\u00b2 + 1<\/strong>
at the point x = c = -2<\/strong> using the alternate form of the derivative<\/strong>.<\/p>\n\n\n\n
Step 1: Use the Alternate Form of the Derivative<\/h3>\n\n\n\n
The alternate form of the derivative is:f\u2032(c)=lim\u2061x\u2192cf(x)\u2212f(c)x\u2212cf'(c) = \\lim_{x \\to c} \\frac{f(x) – f(c)}{x – c}f\u2032(c)=x\u2192clim\u200bx\u2212cf(x)\u2212f(c)\u200b<\/p>\n\n\n\n
Here, f(x)=x3+2×2+1f(x) = x^3 + 2x^2 + 1f(x)=x3+2×2+1 and c=\u22122c = -2c=\u22122.<\/p>\n\n\n\n
Step 2: Compute f(c)f(c)f(c)<\/h3>\n\n\n\n
f(\u22122)=(\u22122)3+2(\u22122)2+1=\u22128+8+1=1f(-2) = (-2)^3 + 2(-2)^2 + 1 = -8 + 8 + 1 = 1f(\u22122)=(\u22122)3+2(\u22122)2+1=\u22128+8+1=1<\/p>\n\n\n\n
Step 3: Plug into the Formula<\/h3>\n\n\n\n
f\u2032(\u22122)=lim\u2061x\u2192\u22122f(x)\u2212f(\u22122)x+2=lim\u2061x\u2192\u22122×3+2×2+1\u22121x+2=lim\u2061x\u2192\u22122×3+2x2x+2f'(-2) = \\lim_{x \\to -2} \\frac{f(x) – f(-2)}{x + 2} = \\lim_{x \\to -2} \\frac{x^3 + 2x^2 + 1 – 1}{x + 2} = \\lim_{x \\to -2} \\frac{x^3 + 2x^2}{x + 2}f\u2032(\u22122)=x\u2192\u22122lim\u200bx+2f(x)\u2212f(\u22122)\u200b=x\u2192\u22122lim\u200bx+2×3+2×2+1\u22121\u200b=x\u2192\u22122lim\u200bx+2×3+2×2\u200b<\/p>\n\n\n\n
Step 4: Simplify the Expression<\/h3>\n\n\n\n
Factor the numerator:x3+2×2=x2(x+2)x^3 + 2x^2 = x^2(x + 2)x3+2×2=x2(x+2)<\/p>\n\n\n\n
So the limit becomes:f\u2032(\u22122)=lim\u2061x\u2192\u22122×2(x+2)x+2f'(-2) = \\lim_{x \\to -2} \\frac{x^2(x + 2)}{x + 2}f\u2032(\u22122)=x\u2192\u22122lim\u200bx+2×2(x+2)\u200b<\/p>\n\n\n\n
Cancel x+2x + 2x+2 (as long as x\u2260\u22122x \\ne -2x\ue020=\u22122):=lim\u2061x\u2192\u22122×2= \\lim_{x \\to -2} x^2=x\u2192\u22122lim\u200bx2=(\u22122)2=4= (-2)^2 = 4=(\u22122)2=4<\/p>\n\n\n\n
Final Answer:<\/h3>\n\n\n\n
f\u2032(\u22122)=4\\boxed{f'(-2) = 4}f\u2032(\u22122)=4\u200b<\/p>\n\n\n\n
\n\n\n\n
Explanation <\/h3>\n\n\n\n
To find the derivative of a function at a point using the alternate form, we apply the definition:f\u2032(c)=lim\u2061x\u2192cf(x)\u2212f(c)x\u2212cf'(c) = \\lim_{x \\to c} \\frac{f(x) – f(c)}{x – c}f\u2032(c)=x\u2192clim\u200bx\u2212cf(x)\u2212f(c)\u200b<\/p>\n\n\n\n
This expression gives the slope of the tangent line to the curve at the point x=cx = cx=c. Instead of using rules like the power rule, this method uses limits to find the derivative from the basic definition.<\/p>\n\n\n\n
In our example, the function is f(x)=x3+2×2+1f(x) = x^3 + 2x^2 + 1f(x)=x3+2×2+1, and we are finding the derivative at c=\u22122c = -2c=\u22122. First, we evaluate f(\u22122)f(-2)f(\u22122) and substitute into the formula. Simplifying the expression gives us a rational function with a removable discontinuity at x=\u22122x = -2x=\u22122. By factoring the numerator, we can cancel out the x+2x + 2x+2 term in both numerator and denominator, which resolves the indeterminate form.<\/p>\n\n\n\n
After canceling, we evaluate the remaining expression x2x^2×2 at x=\u22122x = -2x=\u22122, which results in 444. This is the slope of the tangent line to the curve at the point x=\u22122x = -2x=\u22122.<\/p>\n\n\n\n
This method is valuable for understanding what a derivative truly means: the instantaneous rate of change of a function at a specific point. It also reinforces the concept of limits and how algebraic simplification can help us evaluate them when direct substitution gives an indeterminate form.<\/p>\n\n\n\n
\"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"
f(x) = x^3 + 2x^2 + 1, c = -2 The Correct Answer and Explanation is: We are asked to find the derivative of the functionf(x) = x\u00b3 + 2x\u00b2 + 1at the point x = c = -2 using the alternate form of the derivative. Step 1: Use the Alternate Form of the Derivative […]<\/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-26437","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26437","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=26437"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26437\/revisions"}],"predecessor-version":[{"id":26439,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26437\/revisions\/26439"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=26437"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=26437"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=26437"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}