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

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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

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

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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) =?