# $f(1)=-3$ and $f'(x)\geq7$ how small is $f(5)$?

Here is a test question in my textbook.

Suppose : $f(1)=-3$ and $f'(x)\geq7$ How small $f(5)$ can be possibly :

a)$25$
b)$-21$
c)$28$
e)$31$
f) None of others.

I just have this : because $f'(x)>0$ then $f(x)$ increasing. So, for all $x$ greater than $1$, $f(x)$ will greater than $-3$.

After this conclusion, I don't have any idea.

Because of the above conclusion, I post a,b,c. Because (I think) this problem will not have a fixed solution. So, we should choose which solution is the most approximate.

• @Serkan is $f(x)$ linear, right ? But can we have this conclusion ? (I think this obvious, but maybe there some other function, that after you derive, it will be constant too)
– hqt
Jun 13 '12 at 18:26
• The idea is to think about what the smallest possible value $f(5)$ can be under the conditions given. Jun 13 '12 at 18:27
• Hint: Use the fundamental theorem of calulus and express $f(5)$ in terms of $f(-3)$ and $f'$. Jun 13 '12 at 18:30
• Another hint: if what you know is '$f'(x)\geq 0$ implies $f(x)$ increasing', then consider the function $g(x)=f(x)-7x$. What can you say about $g(1)$, $g'(x)$, and thus $g(5)$? And what does that tell you about $f(5)$? Jun 13 '12 at 18:33

Hint

$$\frac{f(5)-f(1)}{5-1} = f'(c) \geq 7 \,.$$

A previous answer uses an intuitive "let's use a linear function with a minimal derivative" solution.

For a rigorous argument, one can use the mean value theorem (in undergraduate mathematics, this is considered a "basic" theorem). This theorem says that, if $f$ is continuously differentiable*, then there must be an $x\in[1,5]$ (that is, an $x$ between 1 and 5) such that:

$$f'(x)=\frac{f(5)-f(1)}{5-1}$$

First note that I do not know what $x$ is. It is definitely in the range $[1,5]$, but the mean value theorem doesn't tell us what it is.

Also notice that I'm using a strict equality here - mathematicians like that, because we can now solve for $f(5)$ (which is good, to say the least):

$$f(5)=4f'(x)+f(1)$$

But now it is easy! If $f'(x)\geq7$ (for /any/ $x$), then $4f'(x)\geq28$, so $4f'(x)+f(1)\geq28+f(1)=25$, thus answer a) would be correct.

*There are many functions which are not continuously differentiable, but we usually do not consider them. Also, the question implicitly stated that the second derivative exists for all $x$ (namely, for all $x$ it has some value greater than or equal to 7), which is sufficient for the mean value theorem.

• This is the rigorous way of doing it; +1. Jun 13 '12 at 18:49
• Didn't N.S. say the same thing? Jun 13 '12 at 18:54
• @Gigili N.S. did not specify the MVT, so I thought I'd at least clarify that - N.S.'s answer might as well have been an educated guess. Jun 13 '12 at 18:56
• That makes sense, plus one. Jun 13 '12 at 18:57

So you have a line of the form $y=7x+c$ , we take $7$ here because we want the minimum , if you take more then it will be more steep hence you get high value. To get $c$ you have initial condition , put $x=1$ and $y=-3$ to get $c=-10$ . Now you have $y=7x-10$ Put $x=5$ to get $y=25$

• ah. put your solution together with Steven Stadnicki comment to prove f(x) is a linear equation, can solve my question :D
– hqt
Jun 13 '12 at 18:37
• And. this problem is call "how small" make me think about min/max rather than think $f(5)$ is a fixed number :D
– hqt
Jun 13 '12 at 18:38
• Well, because there's an inequality, there are many functions $f$ that satisfy the inequality on the derivative $f'(c) \geq 7$; the key point is to recognize that only one (the linear function) keeps the value of $f(5)$ as small as possible. For instance, $f(x) = x^3+7x-5$ satisfies $f(1) = -3$ and $f'(c) = 3c^2+7 \geq 7$ for all c, but $f(5) = 155$. Jun 13 '12 at 18:48