We are given a set of statements about basic concepts in calculus and asked to determine if they are true or false. The statements cover topics like local extrema, antiderivatives, decreasing functions, differentiability, critical points, and the relationship between a function and its derivative.

AnalysisCalculusDerivativesAntiderivativesLocal ExtremaIncreasing/Decreasing FunctionsDifferentiabilityCritical PointsMean Value Theorem
2025/6/27

1. Problem Description

We are given a set of statements about basic concepts in calculus and asked to determine if they are true or false. The statements cover topics like local extrema, antiderivatives, decreasing functions, differentiability, critical points, and the relationship between a function and its derivative.

2. Solution Steps

6. If $f'(c) = 0$, the function $f$ has a local maximum or minimum at $c$.

This statement is TRUE. A critical point cc where f(c)=0f'(c) = 0 is a candidate for a local maximum or local minimum.

7. Any antiderivative of $f'(x)$ is a function $f(x) + C$ where $C$ is any real number.

This statement is TRUE. The general antiderivative of a function is always defined up to a constant CC.

8. If $f(x)$ is decreasing and $f(x) > 0$ on $I$, then $1/f(x)$ is also decreasing on $I$.

This statement is FALSE. If f(x)f(x) is decreasing and positive, then 1/f(x)1/f(x) is increasing. If f(x)f(x) is decreasing, f(x)<0f'(x) < 0. Let g(x)=1f(x)g(x) = \frac{1}{f(x)}. Then g(x)=f(x)[f(x)]2g'(x) = -\frac{f'(x)}{[f(x)]^2}. Since f(x)<0f'(x) < 0 and [f(x)]2>0[f(x)]^2 > 0, then g(x)>0g'(x) > 0, meaning g(x)g(x) is increasing.

9. If a function $f(x)$ is differentiable at any number $x_0$, then the function $\ln(f(x))$ is also differentiable at $x_0$.

This statement is FALSE. We need f(x0)>0f(x_0) > 0 for ln(f(x0))\ln(f(x_0)) to be defined. If f(x0)0f(x_0) \le 0, then ln(f(x0))\ln(f(x_0)) is not defined (or is not a real number if f(x0)<0f(x_0) < 0), and therefore not differentiable. Even if f(x0)>0f(x_0) > 0, we also require that f(x0)f'(x_0) exists and f(x0)0f(x_0) \ne 0.
1

0. If $f'(a)$ does exist then $a$ is not a critical number of the function $f(x)$.

This statement is FALSE. A critical number aa is defined where either f(a)=0f'(a)=0 or f(a)f'(a) does not exist. Since f(a)f'(a) exists, f(a)f'(a) must be

0. Then $a$ would be a critical number.

1

1. If $f'(x)$ exists and is nonzero for all $x$, then $f(1) \ne f(0)$.

This statement is TRUE. Since f(x)f'(x) exists and is nonzero for all xx, the function is either strictly increasing or strictly decreasing. By the Mean Value Theorem, there exists a cc in (0,1)(0, 1) such that f(c)=f(1)f(0)10=f(1)f(0)f'(c) = \frac{f(1)-f(0)}{1-0} = f(1) - f(0). Since f(c)0f'(c) \ne 0, f(1)f(0)0f(1) - f(0) \ne 0, and therefore f(1)f(0)f(1) \ne f(0).
1

2. If $f$ and $g$ are differentiable and $f(x) \ge g(x)$ for $a < x < b$, then $f'(x) \ge g'(x)$ for $a < x < b$.

This statement is FALSE. For example, let f(x)=x2f(x) = x^2 and g(x)=x2g(x) = -x^2 on the interval (1,1)(-1, 1). Then f(x)g(x)f(x) \ge g(x) on this interval. However, f(x)=2xf'(x) = 2x and g(x)=2xg'(x) = -2x. For x<0x < 0, f(x)<g(x)f'(x) < g'(x), so the statement is false.
1

3. All continuous functions have derivatives.

This statement is FALSE. A classic example is f(x)=xf(x) = |x|, which is continuous everywhere but not differentiable at x=0x = 0.

3. Final Answer

6. True

7. True

8. False

9. False

1

0. False

1

1. True

1

2. False

1

3. False

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