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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.
Aug 26, 2020 · Here is a proposed definition: A function f is said to be concave up on an interval [a, b] if for all x, y ∈ [a, b] with x <y, the line L connecting (x, f(x)) and (y, f(y)) satisfies L(t) ≥ f(t) for all t ∈ (x, y). It is concave down if the final inequality is reversed. Share. Improve this answer.
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.
Consider a function $F(L)=(L-L^2a)T^{L-1}$, where $0<L<\frac{1}{a}$. The constants $a$ and $T$ may take values over $]0,1[$ and $[0.01,0.1]$, respectively. The first derivative of $F$: $\fr...
Jul 25, 2021 · First, we will focus on our function analysis and find intercepts, domain, range, symmetry, asymptotes, and continuity. X Intercepts: f (x) = 0 − 1 3 (x 3 − 3 x + 2) = 0 − 1 3 (x − 1) (x − 1) (x + 2) = 0 x = 1, − 2 (1, 0) (− 2, 0) Y Intercepts: f (0) = − 1 3 (0) 3 + (0) − 2 3 = − 2 3 (0, − 2 3)
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Dec 21, 2020 · Example \(\PageIndex{1}\) Describe the concavity of \( f(x)=x^3-x\). Solution. The first dervative is \( f'(x)=3x^2-1\) and the second is \(f''(x)=6x\). Since \(f''(0)=0\), there is potentially an inflection point at zero.
