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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?
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.
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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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?.
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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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?, a detailed solution 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? has been provided alongside types of 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? theory, EduRev gives you an
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