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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
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.
Dec 21, 2020 · If we are trying to understand the shape of the graph of a function, knowing where it is concave up and concave down helps us to get a more accurate picture. Of particular interest are points at which the concavity changes from up to down or down to up; such points are called inflection points.
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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f(b)(c a) f(a)(c b) + f(c)(b a); which (since c a > 0) holds i. f(b)baf(a) +aaf(c):Take = (c b)=(c a) 2 (0; 1) and verify. that, indeed, b = a + (1 )c. Then the last inequal. ty holds since f is concave. Conversely, the preceding argument shows that if the rst inequality in (1) holds then f is.
