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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.
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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Dec 21, 2020 · The important \(x\)-values at which concavity might switch are \(x=-1\), \(x=0\) and \(x=1\), which split the number line into four intervals as shown in Figure \(\PageIndex{7}\). We determine the concavity on each.
Oct 18, 2018 · Learning Objectives. Use separation of variables to solve a differential equation. Solve applications using separation of variables. We now examine a solution technique for finding exact solutions to a class of differential equations known as separable differential equations.
If \( f''(x) \) is positive on an interval, the graph of \( y=f(x) \) is concave up on that interval. We can say that \(f\) is increasing (or decreasing) at an increasing rate. If \( f''(x) \) is negative on an interval, the graph of \( y=f(x) \) is concave down on that interval.
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May 22, 2016 · We know for single variable function $f(t)$, the necessary and sufficient condition for concavity is $$ f((1-\lambda)x+\lambda y) \ge (1-\lambda)f(x)-\lambda f(y) $$ for every $x$ and $y$ and $0 \le \lambda \le 1$.
