True or false about derivatives

Question

So I have a question on a few of my homework problems mostly because I am ambiguous about the answers and how I thought about them. So...

1) the derivative of f(x)=π^e is f'(x)=eπ^(e-1)

I said this was false because this wasn't a power function, but I don't know how to find the true derivative to disprove it so I'm a little stuck on this one.

2) If f(x) is increasing, then f(x) is increasing.

I said false for this as well because f(x) increasing just means f'(x) is positive but it could be decreasing right? I don't know a function that does this though so I am a little confused. I would say the absolute value of x but that increases and decreases so that wouldn't work.

3) There is no function such that f'(x)=f(x) for all x.

This one I don't even have a guess on, I am confused as to what it is stating.

4) If f(x) is defined for all of x, then f'(x) is defined for all of x.

This I said was false because the function could have a sharp turn or cusp and still be defined but this would make the f'(x) undefined. I think the absolute value of x would work here to disprove this right?

 

Please help me. I am so lost when it comes to derivatives.

Answer 1

(1)  The given function is a constant so its derivative is zero.

 

(2)  If f(x) is increasing, f'(x) is positive.  But the function could be leveling off, for example, converging toward a horizontal asymptote, and f'(x) could be decreasing even as the function is increasing.

 

(3)  False.  Think of f(x) = e^x.  f'(x) = e^x.

 

(4)  Your reasoning about a sharp turn or cusp is correct.  If the function has a derivative at a point, the function is "smooth" at that point.  But you can have a cusp where the derivative when approaching from the left is different from the derivative when approaching from the right, so f'(x) is not well defined even though f(x) is well defined.

Answer 2

Hi Olivia.

 

1) I'm a little confused about this. It looks like you have pi to the power of e. If so, this is a constant. It's roughly equal to 22.4, so its derivative is 0, like with any other constant. This is therefore false.

 

2) I'm assuming you meant to say "If f(x) is increasing, then f'(x) is increasing." If this is what you meant, then this is false. If f'(x) is increasing, that means the derivative of f'(x) is positive, so f''(x) is positive. This isn't the case for any increasing function that is "concave down."

 

Take f(x) = √x for example. This function is always increasing, since its derivative is

 

f'(x) = 1/(2√(x))

 

which is always positive since x is only defined with positive values (x > 0).

 

However, the derivative of f'(x) is

 

f''(x) = -1/(4√(x³))

 

which is always negative, meaning f'(x) is always decreasing. This is the concave down aspect of the function. It's said to be "increasing at a decreasing rate." So no, just because f(x) is increasing, it does not mean f'(x) is increasing. False.

 

3) Sure there is. e^x is its own derivative. This is false.

 

4) This is also false. Take for example

 

f(x) = x^1/3

 

This is the cube root of x, which is defined everywhere. Its derivative, however, is

 

f'(x) = 1/(3*x^2/3)

 

which is not defined for x = 0.

Answer 3

1)

 

π and e are both constant.  Since a constant raised to a constant is a constant, the derivative of a constant is always zero.

 

 

2)

 

f(x) increases when f'(x) increases.  There are times when f'(x) change signs around the same critical point.  A parabola is a good example of this.  If f(x) increases, then f(x) increases best fits a linear function of positive slope.

 

 

3)

 

f'(x) is the slope of the tangent line at any point of a function.  Therefore, f(x)=f'(x) for all values of x.

 

 

4)

 

This is true for all continuous functions.