Question

Find a function f such that f '(x) = 5x3 and the line 40x + y = 0 is tangent to the graph of f.

Answer 1

The function that is a  tangent to the graph of f is $f(x) = (5/4)x^4+60$

What is the concept of integral?

The definite integral can be interpreted as the resulting area of a region. In addition, it is a value in its result, that is, it does not depend on the variable x, which can be exchanged for any other variable without changing the value of the integral.

Knowing that :

• 40x + y = 0 is tangent of f(x)
• slope of the line = -40

therefore, f'(x)|x=a = -40, so only look for a we have:

$5a^3 = -40\\a^3 = -8\\a = -2$

if, x = -2, then y = 80 (since, 40x + y = 0), therefore, f(x) passes through (-2,80), so in the equation:

$f'(x) = 5x^3\\f(x) = (5/4)x^4 + c\\80 = (5/4)(-2)^4 + c\\c = 60$

Therefore,

$f(x) = (5/4)x^4+60$

See more about tangent at brainly.com/question/401236

#SPJ1