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Give an example of a nonconstant function \(f:(-1,1) \rightarrow \mathbb{R}\) such that \(f\) has a local extremum at 0, and 0 is a point of inflection for \(f\).

Short Answer

Expert verified
An example of a nonconstant function \(f(x)\) defined on the interval (-1, 1) with a local extremum and a point of inflection at x = 0 is \(f(x) = x^4\).

Step by step solution

01

Choose an appropriate function

Let's consider the polynomial function \(f(x) = x^4\). This function is non-constant and defined on the open interval (-1,1).
02

Calculate the first and second derivatives of f

We'll now verify if f'(0) = 0 and f''(0) = 0. 1. Calculate the first derivative of f(x): \(f'(x) = \frac{d}{dx}(x^4) = 4x^3\) 2. Calculate the second derivative of f(x): \(f''(x) = \frac{d^2}{dx^2}(x^4) = \frac{d}{dx}(4x^3) = 12x^2\)
03

Check if f'(0) = 0 and f''(0) = 0

Evaluate f'(0) and f''(0) using the derivatives we found in Step 2: 1. \(f'(0) = 4(0)^3 = 0\) 2. \(f''(0) = 12(0)^2 = 0\) As we can see, f'(0) = 0 and f''(0) = 0, so 0 is both a local extremum and a point of inflection for f(x) = x^4. Therefore, the function \(f(x) = x^4\) is an example of a nonconstant function defined on the interval (-1, 1) with a local extremum and a point of inflection at x = 0.

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