2

My Attempt :-

$f'(x)=12x^2+6x-1$ where $f'(x) \ge \frac{-7}{4}$. So (I think) from LMVT we can directly say that $$\frac{f(b)-f(a)}{b-a} \ge \frac{-7}{4}$$ But the answer Given is $$\frac{f(b)-f(a)}{b-a} > \frac{-7}{4}$$

So my Question is Is This Application of LMVT correct ,if yes then why $\frac{-7}{4}$ is excluded from the given range ?

Freelancer
  • 954

2 Answers2

1

The mean value theorem does not tell you that every $c$ has a corresponding $a$ and $b$ with $\frac{f(a)-f(b)}{a-b}=f'(c)$. It tells you that given $a$ and $b$ you can find a $c$, but it doesn't say given $c$ you can find $a$ and $b$. In your case,

$\dfrac{f(a)-f(b)}{a-b}=4(a^2+ab+b^2)+3(a+b)-1=\dfrac{4(2a+2b+1)^2+(4a+1)^2+(4b+1)^2}{8}-\dfrac{7}{4}$.

The minimum of this thus occurs only for $a=b=-\frac{1}{4}$, but this is not allowed because then $a-b$ is $0$ in the denominator. So there is no such $a$ and $b$ that gives you $-\frac{7}{4}$.

Bob Jones
  • 2,245
1

Intuitively, as long as $f'(x)$ attains its minimum at one single point, there won't exist two distinct points where the secant matches that slope.

To formalize, assume $f'(x)$ has a global minimum $m$, then it follows from MVT that:

$$\frac{f(b)-f(a)}{b-a} \ge m \quad \text{for} \;\;\forall a \ne b$$

If $a,b$ points exist where the equality holds, then by MVT for the intervals $(a,\frac{a+b}{2})$ and $(\frac{a+b}{2},b)$:

$$m = \frac{f(b)-f(a)}{b-a} = \frac{1}{2}\left(\frac{f(b)-f(\frac{a+b}{2})}{\frac{b-a}{2}} + \frac{f(\frac{a+b}{2})-f(a)}{\frac{b-a}{2}}\right) \ge \frac{1}{2}(m+m) = m $$

It follows that the middle inequality must be an equality as well, so:

$$m = \frac{f(b)-f(a)}{b-a} = \frac{f(b)-f(\frac{a+b}{2})}{\frac{b-a}{2}} = \frac{f(\frac{a+b}{2}) - f(a)}{\frac{b-a}{2}} $$

From which:

$$f\left(\frac{a+b}{2}\right) = \frac{f(a) + f(b)}{2}$$

It follows that $f(x)$ is both midpoint convex and concave, thus linear by continuity (see for example Function that is both midpoint convex and concave). But the given $f(x)$ is a $3^{rd}$ degree polynomial which is obviously not linear, so such points $a,b$ do not exist. Therefore the strict inequality holds.

Community
  • 1
dxiv
  • 76,497
  • I understood your answer up to part on how $$\frac{1}{2}\left(\frac{f(b)-f(\frac{a+b}{2})}{\frac{b-a}{2}} + \frac{f(\frac{a+b}{2})-f(a)}{\frac{b-a}{2}}\right) \ge m$$ But how did you conclude that the equality $$ \frac{1}{2}\left(\frac{f(b)-f(\frac{a+b}{2})}{\frac{b-a}{2}} + \frac{f(\frac{a+b}{2})-f(a)}{\frac{b-a}{2}}\right) = m$$ will be satisfied ? – Freelancer Nov 16 '16 at 08:05
  • 1
    If you look at the two far ends of that it comes down to $m \ge m$ but of course $m=m$ so the inequality in between must be an equality. – dxiv Nov 16 '16 at 08:06
  • Oh!!..How silly of me to miss that ....Also Does this also prove that for any given polynomial function $f(x)$(degree $\ge 3$) we will never be able to find points such that $f'(c)=\frac{f(b)-f(a)}{b-a}$ where $c$ is a point of maxima or minima of $f'(x)$ ? – Freelancer Nov 16 '16 at 08:15
  • 1
    That's the case for degree $3$ since $f'(x)$ has exactly one extremum. Not necessarily so for higher degrees. – dxiv Nov 16 '16 at 08:20