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Let f be a real valued differentiable function on R such that f(1)=1. If the y-intercept of the tangent at any point P(x,y) on the curve y=f(x) is equal to the cube of the abscissa of P, then the value of f(-3) is equal to?
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Most Upvoted Answer
Let f be a real valued differentiable function on R such that f(1)=1. ...
Analysis:
To find the value of f(-3) using the given condition, we first need to understand the relationship between the curve y=f(x) and the tangent lines at any point P(x,y) on the curve.

Given Conditions:
- f is a real-valued differentiable function on R.
- f(1) = 1.
- The y-intercept of the tangent at any point P(x,y) on the curve y=f(x) is equal to the cube of the abscissa of P.

Strategy:
- We know that the y-intercept of a line is the value of y when x=0.
- Let's consider a point P(x,y) on the curve y=f(x) and find the equation of the tangent line at this point.
- The y-intercept of this tangent line will be (0, 0) based on the given condition.
- Use the equation of the tangent line to determine the value of f(-3).

Calculations:
- Let's consider a point P(x,y) on the curve y=f(x).
- The slope of the tangent line at P(x,y) is f'(x), the derivative of f(x).
- The equation of the tangent line passing through (x,y) is y = f'(x)(x - x) + f(x) = f(x).
- Since the y-intercept of this line is (0, 0) and f(0) = 0, we have 0 = f'(x)(0 - x) + f(x) = -xf'(x) + f(x).
- Given that the y-intercept is the cube of the abscissa, we have 0 = -x^3 + f(x).
- Setting x = -3, we get f(-3) = -(-3)^3 = -27.
Therefore, the value of f(-3) is -27 based on the given conditions and calculations.
Community Answer
Let f be a real valued differentiable function on R such that f(1)=1. ...
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Not differentiable then S 2019 won is an empty set to equals minus 2 minus 1 0 1 2 3 equals minus 2 minus 1 1 2 4 equals minus 2 2 ANS 2 solution FX is not differentiable at negative 2 minus 1 and 2 s equals minus 2 minus 1 0 1 2 qu 21 let F be a differentiable function such than FXX greater than 0 and F14 then 2019 1 exists and equals 4/7 2 exists and equals 4 3 does not exist 4 exists and equals 0 ANS 2 solutions reply equals FX solution of differential equation 22 let's denote the greatest integer less than or equal to 10 spend 2019 equals pi plus 1 11 equals 0 and 1 solution just since it does not exist You 23 2019-102 like four for one ANS 4 solution huge 24 let F equals RB differentiable at CR and F C equals 0 If GX equals FX then at X equals C G is 20191 not differentiable if FC equals 0 to differentiable if FC03 differentiable if FC equals 04 not differentiable ANS 3 solution Q25 is continuous at x equals 0 than the ordered pair PQ is equal to 2019 ANS 3 solution Q26 LED FRRB a continuously differentiable function such that F2 equals 6 and F2 equals 148 2019 1 18/24 3/12/436 ANS 1 solution Q27 2019/1424 square root 2 3/8 square root 248 A for solution Q28 2019 ANS 2 solution huge 29 if the function has to find on is continuous then K is equal to 2019 1 2 2 1 have 3 1 4 1 / square Benz FX is continuous then Q30 Let FX equals 15 minus X 10 X are then the set of all values of X at This is not differentiable is 2019-15 to 10 15 3 5 10 15 24 10 ANS 1 solution Since FX equals 15 minus 10X GX equals FFX equals 15 minus 10 minus 15 minus 10X equals 15 10X 5 then the points were function GX is non-differential or 10 X equals 0 and 10 X equals 5X equals 10 and X 10 equals plus minus 5 X equals 10 and X equals where it's denotes the greatest integer function then 2019 one F is continuous at X equals 4 ANI One solution Q32 if the function is continuous at x equals 5 then the value of A B is 2019 ANS 4 solution function is continuous at x equals 5 LHL equals RHL5 minus pi B plus 3 equals 5 minus pi A plus 1 Q3 and 1914 square root 22 square root 232 square root 244 ANS 1 solution Q34 let FRRB a differential function satisfying F3 plus F2 equals 0 then is equal to 2019 1 1 2 E 1 3 4 E2ANS 1 solution Q35 if FRR is a differentiable function and F2 equals 6 then 2019/124F222F230412F2 ANS 4 solution Using L hospital rule and Live NetSerum we get putting X = 2 2 F2 F2 = 12 F2 F2 = 6 Q36 2019/18/33 seconds 4/3 AMS One solution GX = FX + FX Then in the Interval - 2/2 G is one differentiable at all points two not continuous three not differentiable at two points four not differentiable at one point ANS 4 solution GX is non-differentiable at x equals 1 GX is not differentiable at one point Q38 let KB the set of all real values of x were the function fx = sin x minus x plus 2x minus pi cos x is not differentiable Then the set k is equal to 2019 empty set 2 pi 3 0 4 0 pi ANS 1 solution Then function f x is differentiable for l Related: JEE Main Previous year questions (2016-20): Limits, Continuity and Differentiability?

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Let f be a real valued differentiable function on R such that f(1)=1. If the y-intercept of the tangent at any point P(x,y) on the curve y=f(x) is equal to the cube of the abscissa of P, then the value of f(-3) is equal to?
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Let f be a real valued differentiable function on R such that f(1)=1. If the y-intercept of the tangent at any point P(x,y) on the curve y=f(x) is equal to the cube of the abscissa of P, then the value of f(-3) is equal to? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Let f be a real valued differentiable function on R such that f(1)=1. If the y-intercept of the tangent at any point P(x,y) on the curve y=f(x) is equal to the cube of the abscissa of P, then the value of f(-3) is equal to? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let f be a real valued differentiable function on R such that f(1)=1. If the y-intercept of the tangent at any point P(x,y) on the curve y=f(x) is equal to the cube of the abscissa of P, then the value of f(-3) is equal to?.
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