These are the local extrema for f (x) = x4 −4x f (x) = x 4 - 4 x. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor.

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2024年8月12日 · Example 4 : Determine the critical points of g(x) = x⁴ – 4x³ + 6x² and classify them as local maxima, local minima, or neither. Solution: g'(x) = 4x³ – 12x² + 12x = 4x(x² – 3x + 3)

Since [latex]f^{\prime}(x)=x^2-5x+4=(x-4)(x-1)[/latex], the critical points are [latex]x=1[/latex] and [latex]x=4[/latex]. From the graph of [latex]f[/latex] in (Figure), we see that [latex]f[/latex] has a local maximum at [latex]x=1[/latex] and a local minimum at [latex]x=4[/latex].

2016年6月28日 · How do you find the local extrema for f (x) = x4 − 4x³? There is a minimum for x = 3. First of all we search the zeros of the derivative. f '(x) = d dx (x4 −4x3) = 4x3 −12x2. f '(x) = 0 → 4x3 − 12x2 = 0. One solution is of course x = 0 the second is …

2024年12月16日 · Find also the local maximum and the local minimum values, as the case may be: (viii) f(𝑥) = 𝑥√(1−𝑥), 𝑥 > 0= √(1−𝑥) + 1/(2√(1 − 𝑥)) (0 −1) . 𝑥 = √(1−𝑥) – 𝑥/(2√(1 − 𝑥)) = (2(√(1 − 𝑥) )^2− 𝑥)/(2√(1 − 𝑥)) =

Let f be continuous on an open interval (a,b) that contains a critical x-value. 1) If f'(x) > 0 for all x on (a,c) and f'(x)<0 for all x on (c,b), then f(c) is a local maximum value. 2) If f'(x) < 0 for all x on (a,c) and f'(x)>0 for all x on (c,b), then f(c) is a local maximum value. 3) If f'(x) has the same sign on both sides of c, then f(c ...

These are the local extrema for f (x) = x4 ln(x) f ( x) = x 4 ln ( x). Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor.

When x = −3, the derivative changes from being negative to being positive, so f has a local minimum. When x = 1/3, the derivative changes from being positive to being negative, so f has a local maximum. f(x) = x4 − 4x2 + 3. f′(x) = 4x3 − 8x = 4x(x2 − 2) = 4x(x − √2)(x + √2). The solutions of f′(x) = 0 are thus x = 0 and x = ±√2.

2019年7月1日 · If "local minimum value" means the function value in the local minimum, b) is correct. If it means the number of local minima, d) is correct.