Find the differential of f(x)=y where y=3x^2+2x+4. Hence find the coordinates of the minimum point of f(x)

The differential of f(x) is f'(x)=6x+2 because we apply this basic formula: the differential of kxⁿ = knx^n-1 where k is a constant.This is a positive quadratic, so it has a U shape that is above the x-axis (I would draw a sketch of the graph at this point). This means it has only got one point where the gradient is zero. This must be the minimum point of f(x) (I would demonstrate this on the sketch of the graph).Therefore since f'(x) tells us the gradient of f(x), the minimum point will be where f'(x)=0, so we can say 6x+2=0.Solving for this gives x=-1/3. To find the y value for the minimum point we calculate f(-1/3) by substituting x=-1/3 back into the original equation y=3x²+2x+4. This gives a value of 11/3.Hence the minimum point of f(x) is (-1/3, 11/3)