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State the first derivative test for critical points. Use concavity and inflection points to explain how the sign of the second derivative affects the shape of a function’s graph. Explain the concavity test for a function over an open interval.
Dec 21, 2020 · The number line in Figure \(\PageIndex{5}\) illustrates the process of determining concavity; Figure \(\PageIndex{6}\) shows a graph of \(f\) and \(f''\), confirming our results. Notice how \(f\) is concave down precisely when \(f''(x)<0\) and concave up when \(f''(x)>0\). Figure \(\PageIndex{6}\): A graph of \(f(x)\) used in Example ...
Review your knowledge of concavity of functions and how we use differential calculus to analyze it.
- You have two functions f and g, where f''>0 and g''>0. If you take the second derivative of f+g, you get f''+g'', which is positive. So their sum i...
- In order for 𝑓(𝑥) to be concave up, in some interval, 𝑓 ''(𝑥) has to be greater than or equal to 0 (i.e. non-negative) for all 𝑥 in that inter...
- In general, no. Points where f"(x)= 0 are specifically called inflection points. But in the example provided, -2 is also a critical point (Observe...
- Product rule: d/dx[f(x)*g(x)] = f'(x)g(x) + f(x)g'(x) f(x) = (x-1)³ and g(x) = x
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Use concavity and inflection points to explain how the sign of the second derivative affects the shape of a function’s graph. Explain the concavity test for a function over an open interval. Explain the relationship between a function and its first and second derivatives.
How do you describe the concavity of the graph and find the points of inflection (if any) for #f(x) = x^3 - 3x + 2#?
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If f ′ (x) is positive on an interval, the graph of y = f(x) is increasing on that interval. If f ′ (x) is negative on an interval, the graph of y = f(x) is decreasing on that interval. The second derivative tells us if a function is concave up or concave down.
