# True or False

#### Jkacson

##### New member
So for number 18 on here I'm not sure how to even figure out if they're true or not

#### Dr.Peterson

##### Elite Member
First, think about what theorems you have learned that might apply. If there is one, carefully check its conditions -- for example, if it requires continuity but nothing is said in the problem about that, be suspicious. What you're looking for here is a reason it might be true, together with reasons it might be a sort of trick question.

If you don't find a theorem that applies, look for a counterexample. Think about functions that might not be the first ones you'd think of: If it doesn't say f is continuous, pick a discontinuous function, and so on.

The best way for us to help you would be if you make some attempt at answering each part and tell us your thinking - even if it's mostly a guess. Then we can suggest ideas you might be missing. (The exercise will be most beneficial to you if it makes you think a lot! That's the whole point, so we won't give away the answer before you've had a chance to learn from it!)

#### pka

##### Elite Member
View attachment 11721So for number 18 on here I'm not sure how to even figure out if they're true or not
Except for c), the others should be simple.
a) Look here
c) Use the mean value theorem.
d) $$\displaystyle f(x)=x^3,~x=0$$
e) use chain rule.

#### Jkacson

##### New member
Okay so for a) False because a symmetric graph across the y axis will have 2 points equal to each other
b) false because f(x) and g(x) is not equal to (f*g)(x)
c) I looked over the mean value, but I'm still not understanding how to implement it
d) I got true because critical points occur at the first derivative equal to zero and critical points are where local extrema is
e) I ended getting false from applying the chain rule
I'm also not very good at calculus so I wouldn't be surprised if I'm still doing these wrong

#### Dr.Peterson

##### Elite Member
Okay so for a) False because a symmetric graph across the y axis will have 2 points equal to each other
b) false because f(x) and g(x) is not equal to (f*g)(x)
c) I looked over the mean value, but I'm still not understanding how to implement it
d) I got true because critical points occur at the first derivative equal to zero and critical points are where local extrema is
e) I ended getting false from applying the chain rule
I'm also not very good at calculus so I wouldn't be surprised if I'm still doing these wrong
(a) You haven't described your counterexample very well, but you may be thinking about the right thing. Can you give a specific example of f?
(b) Again, give a specific example. I don't even know what you mean by "f(x) and g(x) is not equal to (f*g)(x)".
(c) Suppose that f(a) = f(b). What does the MVT tell you?
(d) Are you sure that every critical point is a local extremum? Check what your book says about that. There's a big difference between saying "every extremum is a critical point" and "every critical point is an extremum".
(e) Can you show the details of what you did?

#### topsquark

##### Full Member
You misspelled MTV, Doc. I get all my Math from Daria.

-Dan

#### Jomo

##### Elite Member
Okay so for a) False because a symmetric graph across the y axis will have 2 points equal to each other
b) false because f(x) and g(x) is not equal to (f*g)(x)
c) I looked over the mean value, but I'm still not understanding how to implement it
d) I got true because critical points occur at the first derivative equal to zero and critical points are where local extrema is
e) I ended getting false from applying the chain rule
I'm also not very good at calculus so I wouldn't be surprised if I'm still doing these wrong
For (a) you do not need symmetry. Must a function as described in part (a) have an absolute maximum point? Can it have more than one? In my opinion instead of thinking about this you should just draw a continuous function from a to b (do not lift your pen) and see if it has an absolute maximum value. If it has more than one absolute maximum then you should know the answer to the question. If your graph has only one absolute max than ask yourself if you can modify it so it has two or more absolute extremes.

b) Are you saying that f(x) = g(x) = (f*g)(x) can never happen? What if f(x) = g(x) = x, then (f*g)(x) = x AND if f(x) = g(x) = 7, then (f*g)(x) = 7. But for your question move away from that. We know that g(a) is a rel max, say its 104. What if anything do we know about f(104)? Note that (f*g)(a) = f(g(a)) = f(104).

c) Personally I would not use the MVT. What does it mean if a$$\displaystyle \neq$$b, then f(a) $$\displaystyle \neq$$ f(b), ie if a and b are different then f(a) and f(b) must be different? Think carefully about what does that mean (hint: does f(x) = x2 have this property?) and whether or not ALL derivative functions have that property

d) Look at the graph of f(x) = x3 and find all solutions to f' (x) = 0

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#### pka

##### Elite Member
View attachment 11721So for number 18 on here I'm not sure how to even figure out if they're true or not
c) Suppose that $$\displaystyle a<b$$ having the property that $$\displaystyle f(a)=f(b)$$.
By the mean value theorem, $$\displaystyle \exists c\in(a,b)$$ having the property that $$\displaystyle f'(c)=\frac{f(b)-f(a)}{b-a}=0.$$
That is contrary to the given conditions on $$\displaystyle f$$.

d) This has already been posted, but to reinforce it. Inflection points are critical points but by almost universal agreement they are not local extrema. That is true of $$\displaystyle f(x)=x^3$$ at $$\displaystyle x=0$$.

e) Given that $$\displaystyle f(-x)=f(x)$$ (i.e. $$\displaystyle f$$ is an even function) the it follows that:
\displaystyle \begin{align*}f(x)&=f(-x) \\f'(x)&=-f'(-x)\\-f'(x)&=f'(-x) \end{align*}
Which shows $$\displaystyle f'$$ is an odd function.