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

Answer:

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

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) =?