Question

Given f '(x) = (x - 4)(4 - 2x), find the x-coordinate for the relative maximum on the graph of f(x)
A)4
B)3
c)2
D)None of these

Answer 1

What will be true about the value of \(f'(x)\) at a relative maximum?

Answer 2

It will be (3,2)?

Answer 3

Is that an answer to my question, or to the problem?

Answer 4

To your question

Answer 5

so you're saying that for any function, the value of \(f'(x)\) at a relative maximum is (3,2)? :-)

Answer 6

I'm asking a general question here...what do we know about the first derivative at a local maximum?

Answer 7

Oh I thought your question was about my specific question

Answer 8

we will use it to get an answer to your specific question.

Answer 9

Well it changes sign at the relative maximum, right?

Answer 10

the first derivative is the instantaneous slope of the tangent to the function. so at a local maximum or a local minimum, it will be equal to 0.

Answer 11

So how do I apply it to this question?

Answer 12

where are the points where \(f'(x) = (x-4)(4-2x) = 0\)?

Answer 13

Oh... Yeah that makes sense, so the answer would be 2 right? Thank you a lot for your help!

Answer 14

not so fast...

is that the only place where \(f'(x) = (x-4)(4-2x) = 0\)?

Answer 15

No, there are two points where it equals 0

Answer 16

by using the zero product rule,
\[(x-4)(4-2x) = 0\]has solutions at \[(x-4) = 0 \rightarrow x = 4\]\[(4-2x) = 0\rightarrow x = 2\]

We just know (so far) that \(x = 4\) and \(x=2\) are either local maximum or local minimum points on the graph of \(f(x\).

Answer 17

How do you propose to figure out which one is the local maximum?

Answer 18

I don't really know. What do I do? :(

Answer 19

Whapalmer!

Answer 20

do you know how to integrate a polynomial?

Answer 21

if you do, you could integrate \(f'(x)\) and evaluate the result (which would be \(f(x) + C\) where C is an arbitrary constant) at both values of \(x\) and see which one is the local maximum and which is the local minimum. That would one approach.

Another would be to look at the value of \(f'(x)\) between the two inflection points (points where \(f'(x) = 0\)). If that value is positive, that means the slope of the tangent to \(f(x)\) is also positive. What would imply about which point was the local maximum?

Answer 22

is it 4?

Answer 23

Well, I'll tell you if it is or not, if you tell me why you think 4 is the answer...

Answer 24

Let's consider the value of \(f'(1)\). \(f'(x) = (x-4)(4-2x)\) so \(f'(1) = (1-4)(4-2(1)) = -6\) which means that the function is decreasing at \(x = 1\).

Now let's consider \(f'(3) = (3-4)(4-2(3)) = 2\) which means the function is increasing at \(x = 3\).

And finally \(f'(5) = (5-4)(4-2(5)) = -6\) so the function is again decreasing at \(x=5\).
\[\begin{array}{cc}
x &f(x)\text{ is increasing/decreasing}\\
\hline\\1 & \text{decreasing}\\
2&\text{inflection point}\\
3&\text{increasing}\\
4&\text{inflection point}\\
5&\text{decreasing}

\end{array}
\]

Hopefully, from that table you can visualize that \(f(2\) is a local minimum, and \(f(4)\) is a local maximum.