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Find a function f such that f '(x) = 3x^3 and the line 81x + y = 0 is tangent to the graph of f. f(x) =?

Please help me solve this apparently easy calculus question!! I Will be very thankful!! Thank you

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The Best answer for Find a function f such that f '(x) = 3x^3 and the line 81x + y = 0 is tangent to the graph of f. f(x) =?

f '(x) = 3x^3 = -81 => x = -3

y - f(-3) = -81(x + 3)

y = -81x - 243 + f(-3)

f(-3) - 243 = 0 => f(-3) = 243

f(x) = 3/4*x^4 + C

243 = 243/4 + C => C = 182.25

f(x) = 3/4*x^4 + 182.25

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If f'(x) = 3x^3 then f(x) = (3/4)x^4 + c, for some value of c. Since 81x + y = 0 is a tangent then the gradient of this line = -81 = f'(x) = 3x^3 (the dash on f'(x) isn't showing up very well). This means that 3x^3 = - 81, so x^3 = -27, so x = -3. Thereforethe point of contact (since it lies on the line) is

(-3,243) and this point is also on the curve, so (3/4)(-3)^4 +c = 243; this reduces to (243/4) + c = 243

so c = 243 - 243/4 = (3/4)243 , so the equation of the curve is y = (3/4)x + (3/4)243

which might be written y = (3/4)(x + 243) or 4y = 3(x + 243) or 3x + 729

This is the Correct Answer for Find a function f such that f '(x) = 3x^3 and the line 81x + y = 0 is tangent to the graph of f. f(x) =?

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f '(x) = 3x^3

81 x + y = 0

so y = 81 x

y' = 81

Let 3x^3 = 81

x^3 = 27

x = 3

when x = 3 ,

81x + y = 0

y = -81x = -81(3) = -243

ie ( (3 , -243)

f ' (x) = 3x^3

so f(x) = 3/4 x^4 + c

-243 = 3/4 (3)^4 +c

-243 = 3/4 *81 +c

c = -243 - 243 / 4

= - 972/4 - 243/4

= -1215/ 4

so f(x) = 3/4 x^4 - 1215/4

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anti derivative is (3x^4)/4 +C.

y=-81x.

m=-81

-81=3x^3

=-3

y=-81(3) =243

243=3/4(-3)^4 +C

C=729/4

The final answer of the problem is f(x)=(3x^4)/4 +729/4

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