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

Solution:

Let's integrate the function f’(x) to get the original function.

Step 1: Integrate f'(x) = 5x³

∫5x³ dx = 5x⁴/4 + C.

Step 2: To find the slope of tangent line y = -5x,

Differentiate both sides with respect to ‘x’.

dy/dx = f'(x) = -5

Step 3: Let us find the x and y coordinates of the tangent

Find the x coordinate:

f'(x) = 5x³ = -5

x³ = -1

x = -1

Find the y coordinate:

f(-1) = 5(-1)⁴/4 + C

y = 5/4 + C

Step 4: Let's find C to get the original function.

We have m = -5 and the tangent line passing through(-1, 5/4+C)

The equation of the tangent line passing through the points (x\(_1\),y\(_1\)) is given as y-y\(_1\) = m ( x- x\(_1\))

y - (5/4 + C) = - 5(x + 1)

y = - 5x - 15/4 + C.

So -15/4 + C = 0,

C = 15/4

Thus the required function is f(x) = 5x/4 + C.

f(x) = 5x/4 + 15/4

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

Summary:

The function f(x) is 5x⁴/4 + 15/4 when the line 5x + y = 0 is tangent to the graph of f.