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# Justification using second derivative: maximum point

Given that the derivative of a function is zero, we can justify whether the function has a relative maximum point by looking at the second derivative.

## Want to join the conversation?

• I don't think that this was an accurate justification because h''(-5) is also negative, does that mean that h(5) is the relative maximum? NO!!
• True, but h'(5) isn't zero.

The justification is not just that h''(x) is negative; it has to be h'(x) = 0 AND h''(x) is negative.
And we're told in the question that h'(-4) = 0.
• Why the graph of the function concaves downward when the second derivative is negative?
• If a function's second derivative is negative, then its slope is decreasing. This is equivalent to saying that a function is concave downward.

Remember: The first derivative gives the rate of change (slope) of the function, while the second derivative gives the rate of change of the first derivative. This means that the second derivative tells us how the slope is changing.
• How is it possible when f''(x) is negative that f'(x) is 0, shouldn't it be decreasing? am I missing something? [h''(-4) is negative, h'(-4)=0]
(1 vote)
• f''(x)<0 means the function is concave down, and f'(x)=0 means it's neither increasing nor decreasing at that point.

Consider f(x)=-x^2 at x=0:
f'(x)=-2x, so f'(0)=0, and f''(x)=-2.
• Does it matter at all that the second derivative of the function has a local extremum at the same "`x`" value as the function's local extremum?