1e:I[80107,["/_next/static/chunks/235ad55241b6443f.js","/_next/static/chunks/8337ab1f037b6654.js","/_next/static/chunks/c9e933b5fd1a0afc.js","/_next/static/chunks/848a330faa2461ac.js","/_next/static/chunks/9f2942e368a262d2.js","/_next/static/chunks/92f9e11244990470.js","/_next/static/chunks/a8b867fbb1dfca1a.js","/_next/static/chunks/1100f0390a901fa7.js","/_next/static/chunks/bbe6bd9d3c7b0355.js","/_next/static/chunks/a82195c3e5de7dd4.js"],"default"] 1f:I[76763,["/_next/static/chunks/235ad55241b6443f.js","/_next/static/chunks/8337ab1f037b6654.js","/_next/static/chunks/c9e933b5fd1a0afc.js","/_next/static/chunks/848a330faa2461ac.js","/_next/static/chunks/9f2942e368a262d2.js","/_next/static/chunks/92f9e11244990470.js","/_next/static/chunks/a8b867fbb1dfca1a.js","/_next/static/chunks/1100f0390a901fa7.js","/_next/static/chunks/bbe6bd9d3c7b0355.js","/_next/static/chunks/a82195c3e5de7dd4.js"],"default"] 1d:T449,{"@context":"https://schema.org","@graph":[{"@type":"BreadcrumbList","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https://www.divinejee.com"},{"@type":"ListItem","position":4,"name":"Let ƒx be a polynomial of degree 3 such that ƒ 1 10 ƒ1 6 ƒx ","item":"https://www.divinejee.com/question/let-fx-be-a-polynomial-of-degree-3-such-that-f-1-10-f1-6-fx-has-a-critical-point-at-x-1-and-fx"}]},{"@type":"QAPage","mainEntity":{"@type":"Question","name":"Question Let ƒx be a polynomial of degree 3 such that ƒ 1 10 ƒ1 6 ƒx on Topic","text":"Let ƒ(x) be a polynomial of degree 3 such that\nƒ(–1) = 10, ƒ(1) = –6, ƒ(x) has a critical point\nat x = –1 and ƒ'(x) has a critical point at x = 1.\nThe...","answerCount":1,"upvoteCount":0,"datePublished":"2026-01-16","author":{"@type":"Organization","name":"DivineJEE"},"acceptedAnswer":{"@type":"Answer","text":"Check the detailed solution on the page.","upvoteCount":0,"url":"https://www.divinejee.com/question/let-fx-be-a-polynomial-of-degree-3-such-that-f-1-10-f1-6-fx-has-a-critical-point-at-x-1-and-fx"}}}]}20:T551,Let f(x) = ax³ + bx² + cx + d

Given f(-1) = 10, f(1) = -6

$$ \therefore $$ -a + b - c + d = 10 ....(i)

and a + b + c + d = -6 ......(ii)

adding (i) + (ii)

2(b + d) = 4

$$ \Rightarrow $$ b + d = 2 ....(iii)

f'(x) = 3ax² + 2bx + c

Given f'(-1) = 0

$$ \Rightarrow $$ 3a - 2b + c = 0 .....(iv)

f"(x) = 6ax + 2b

Given f"(1) = 0

$$ \therefore $$ 6a + 2b = 0 ....(v)

$$ \Rightarrow $$ b = -3a

adding (iv) + (v), we get

9a + c = 0 ....(vi)

$$ \Rightarrow $$ $$9\left( {{{ - b} \over 3}} \right)$$ + c = 0

$$ \Rightarrow $$ c = 3b

f(x) = $${{{ - b} \over 3}{x^3}}$$ + bx² + 3bx + (2 - b)

$$ \Rightarrow $$ f'(x) = -bx² + 2bx + 3b

= -b(x² - 2x - 3)

At maxima and minima f'(x) = 0

$$ \therefore $$ (x² - 2x - 3) = 0

$$ \Rightarrow $$ (x - 3) (x + 1) = 0

x = 3, -1

As a + b + c + d = -6

$$ \Rightarrow $$ $${{{ - b} \over 3}}$$ + b + 3b + 2 - b = -6

$$ \Rightarrow $$ b = -3

$$ \therefore $$ f'(x) = 3(x² - 2x - 3)

$$ \Rightarrow $$ f''(x) = 3(2x - 2)

At x = 3, f''(x) = 3(2.3 - 2) = 12 > 0

$$ \therefore $$ Minima at x = 3.