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We know that the slope of a curve y = f(x) at the point (x, y) is given as dy/dx.
Now, we are given that dy/dx = 6x^2. We will solve this equation by integrating both sides with respect to x:
=> y = Int [6x^2 dx]
= 2x^3 +C
Now to identify the particular solution whose graph passes through the point (4, 8), we find the value of C for which y= 8 when x = 4
y = 2x^3 +C
=> 8 = 2*4^3 +C
=> C = -128 + 8
=> C= -120
Therefore the curve we are looking for is y = 2x^3 – 120.
Given the function of the slope is f(x) = 6x^2
Then, we know that the curve that has a slope of 6x^2 is integral f(x).
==> Let the curve be F(x) such that:
F(x) = intg f(x) dx
==> F(x) = intg (6x^2 ) dx
= 6x^3/ 3 + C
==> F(x) = 2x^3 + C
But the curve passes through the point ( 4, 8).
Then the point (4,8) should verify the equation of the curve.
==> F(4) = 8
==> 2(4^3) + C = 8
==> 128 + C = 8
==> B = -120.
==> F(x) = 2x^3 - 120
the slope of a curve is given by f'(x).
Therfore f'(x) = 6x^2.
Therefore the curve f(x) = Int f'(x)dx .
f(x) = Int 6x^2 dx = (1/2+1)6x^(2+1) +C = (6/3) x^3 +C.
f(x) = 2x^3+C.
Since f(x) passes therough (4,8) ,
f(4) = 8 , or 2*4^3+C = 8. So 128+C = 8, Or C = 8-128 = -120.
Therefore we rewite with C = -120:
f(x) = 2x^3-128.
Therefore the required curve is f(x ) = 2x^3 -120.
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