Prove f'(0) = 0: An Even Function Homework Solution

  • Thread starter Glissando
  • Start date
In summary, we are trying to prove that for an even function f(x) with a derivative at all points, f'(0) = 0. We start with the equation f(x) = f(-x), which is true for all even functions. Then, using the chain rule, we differentiate both sides to get f'(x) = f'(-x) * (-1). Since f'(x) and f'(-x) are equal, we can rewrite this as f'(x) = -f'(x). Therefore, f'(0) must equal 0 since f'(0) = -f'(-0) and f'(-0) is also equal to f'(0). This proves that f'(
  • #1
Glissando
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Homework Statement


If f(x) is an even function and f'(x) exists for all x, prove that f'(0) = 0. (Hint: Start with an equation that is true for all even functions and differentiate both sides with respect to x.)


Homework Equations


Equation true for all even functions: f(x) = f(-x)


The Attempt at a Solution



f(x) = f(-x)

f(d/dx (x)) = f(d/dx (-x))

f(1) = f(-1)

I'm not sure if I have the notation correct when differentiating...or I could have done something like this:

f(x) = f(-x)

f' (x) = f' (-x)

f'(0) = f'(-0)

Then I'm not too sure as to what to do from there ):

Thanks for all the help <3
 
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  • #2
The derivative of f(x) is f'(x), sure. The derivative of f(-x) isn't equal to f'(-x). You need to use the chain rule to differentiate f(-x).
 
  • #3
Dick said:
The derivative of f(x) is f'(x), sure. The derivative of f(-x) isn't equal to f'(-x). You need to use the chain rule to differentiate f(-x).

But I'm taking the derivative with respect to x... ): why do I need to use the chain rule?
 
  • #4
Glissando said:
But I'm taking the derivative with respect to x... ): why do I need to use the chain rule?

Because f(-x) has the form f(g(x)) where g(x)=(-x). Doesn't it?
 
  • #5
Dick said:
Because f(-x) has the form f(g(x)) where g(x)=(-x). Doesn't it?

IF I am doing this right...

f(x) = f(-x)

f' (x) = f'(-x) + f(-1) derivative of outer function times inner function plus outside function times derivative of inside function o_O?

Thank you for your patience (:!
 
  • #6
Glissando said:
IF I am doing this right...

f(x) = f(-x)

f' (x) = f'(-x) + f(-1) derivative of outer function times inner function plus outside function times derivative of inside function o_O?

Thank you for your patience (:!
No, I think you're confusing with the product rule.
 
  • #7
Glissando said:
IF I am doing this right...

f(x) = f(-x)

f' (x) = f'(-x) + f(-1) derivative of outer function times inner function plus outside function times derivative of inside function o_O?

Thank you for your patience (:!

d/dx of f(g(x)) is f'(g(x))*g'(x), right? Isn't that the chain rule? If you put g(x)=(-x) what do you get for the derivative of f(-x)??
 
  • #8
Dick said:
d/dx of f(g(x)) is f'(g(x))*g'(x), right? Isn't that the chain rule? If you put g(x)=(-x) what do you get for the derivative of f(-x)??

Goodness I hope I'm doing it right this time, thanks for bearing with me.

f'(x) = f'(g(x)) * g'(x)

f'(x) = f'(-x) * (-1)

f'(x) = -f'(-x)...= f'(x)?

God I'm feeling so dumb right now ):!
 
  • #9
Glissando said:
Goodness I hope I'm doing it right this time, thanks for bearing with me.

f'(x) = f'(g(x)) * g'(x)

f'(x) = f'(-x) * (-1)

f'(x) = -f'(-x)...= f'(x)?

God I'm feeling so dumb right now ):!

You shouldn't feel dumb now that you are getting it right. Sure, f'(x)=(-f'(-x)). Put x=0 and tell me what f'(0) must be.
 
  • #10
Dick said:
You shouldn't feel dumb now that you are getting it right. Sure, f'(x)=(-f'(-x)). Put x=0 and tell me what f'(0) must be.

God you're good <3

Thank you so much (:!
 
  • #11
Of course, he is!
 

FAQ: Prove f'(0) = 0: An Even Function Homework Solution

What does it mean for a function to be even?

An even function is a type of function in which for every input value x, the output value is the same as the output value for -x. In other words, the function is symmetrical about the y-axis.

How do you prove that a function is even?

To prove that a function is even, you need to show that f(x) = f(-x) for all values of x. This can be done by substituting -x into the function and simplifying the equation to show that it is equal to f(x).

Why is it important to prove that f'(0) = 0 for an even function?

Proving that f'(0) = 0 for an even function is important because it is a property of even functions. It confirms that the function is symmetrical about the y-axis, and it also allows for easier calculations and simplifications when working with the function.

Can a function be both even and odd?

No, a function cannot be both even and odd. An even function is symmetrical about the y-axis, while an odd function is anti-symmetrical about the origin. Therefore, a function cannot exhibit both of these properties at the same time.

What are some real-life examples of even functions?

Some examples of even functions in real life include the cosine function (cos x) which represents the relationship between the adjacent and hypotenuse sides of a right triangle, and the area of a circle (πr^2) which is symmetrical about the y-axis.

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