If f(x) = 7x2 + 5, then 0 is the critical point. As a result, the interval f grows when f is positive.
Although this may appear like a frivolous issue—t = 0 is designated as a key point in each case—it is occasionally crucial to understand why is it that a position is indeed a critical point. In fact, we'll observe a fact which only holds true at crucial sites where the derivative equals zero in a few sections.
f is maximal because we replace x = 0 well at derivatives at the crucial point. After calculating the derivative, equating x to zero, and arriving at 0, we arrive at the critical point, which is at this position.
The interval just on left of the crucial moment is - infinity f 0 on the this interval increasing whilst also f is negative on the right of the critical point 0 >= f +infinity on this interval f is rising whilst also f is positive. This indicates that f is maximum since at the critical point people substitute x = 0 at the variant we have 5 > 0.
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