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Find sum of all possible values of the parameter $b$ if the difference between the largest and smallest values of the function $f(x)=x^2-2bx+1$ in the segment $[0,1]$ is $4$.

I found that the smallest value of $f(x)=x^2-2bx+1$ is $1-b^2$
But i do not know what will be the largest value of the quadratic expression,whether it is at $x=0$ or $x=1$.
Please help me.Thanks.

Vinod Kumar Punia
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2 Answers2

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We have $f(x)=(x-b)^2+1-b^2$. Now let us separate it into four cases :

Case 1 : For $b\lt 0$, we have $$4=f(1)-f(0)\Rightarrow b=-\frac 32$$ which is sufficient.

Case 2 : For $0\le b\le \frac 12$, we have $$4=f(1)-f(b)\Rightarrow b=3,-1$$ which are not sufficient.

Case 3 : For $\frac 12\lt b\le 1$, we have $$4=f(0)-f(b)\Rightarrow b=\pm 2$$ which are not sufficient.

Case 4 : For $b\gt 1$, we have $$4=f(0)-f(1)\Rightarrow b=\frac 52$$ which is sufficient.

Hence, the answer is $-\frac 32+\frac 52=\color{red}{1}$.

mathlove
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  • Why did the values obtained in case $(2)$ and $(3)$ were not added,what does it mean to be not sufficient?@mathlove – Vinod Kumar Punia Dec 08 '15 at 10:24
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    @VinodKumarPunia: For the case 2, we get $b=3,-1$. But these don't satisfy $0\le b\le \frac 12$. So, these are not sufficient. – mathlove Dec 08 '15 at 10:26
  • How did you identify we have to divide the parameter into four cases in this manner only.I mean from 0 to $\frac{1}{2}$ and from $\frac{1}{2}$ to $1$.How does this $\frac{1}{2}$ click to the mind of question solver. – Vinod Kumar Punia Dec 08 '15 at 10:30
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    @VinodKumarPunia: Because $1/2$ is the midpoint of $0$ and $1$. We have a parabola whose axis is $x=b$. As you see, if $0\le b\le 1/2$, the largest value is $f(1)$, but if $1/2\le b\le 1$, the largest value is $f(0)$. So, we need to separate it into four cases. Drawing the graph of the parabola should help. – mathlove Dec 08 '15 at 10:34
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Since $f(x) = (x-b)^2 + (1 - b^2)$, the vertex will be at $(b, 1-b^2)$.

Also $f(1) = 2-2b$ and $f(0) = 1$.

When the vertex is at or to the left of $x=0$, then $f(x)$ is increasing on $[0, 1]$. So $$ \max_{x \in [0,1]}f(x) - \min_{x \in [0,1]}f(x) = f(1) - f(0) = 1-2b$$ and $1-2b=4$ when $b = -\frac 32$

When the vertex is at or to the right of $x=1$, then $f(x)$ is decreasing on $[0, 1]$. So $$ \max_{x \in [0,1]}f(x) - \min_{x \in [0,1]}f(x) = f(0) - f(1) = 2b-1$$ and $2b-1=4$ when $b = \frac 52$

When the x-coordinate of the vertex is in the interval $[0,\frac 12]$ then

$\begin{align} \max_{x \in [0,1]}f(x) - \min_{x \in [0,1]}f(x) &= \max\{f(0),f(1)\} - (1-b^2)\\ &= f(1) - (1-b^2)\\ &= (b-1)^2 \end{align}$

This does not equal $4$ when the x-coordinate of the vertex, b, is in the interval $[0,\frac 12]$

Finally, when the x-coordinate of the vertex is in the interval $(\frac 12, 1]$ then

$\begin{align} \max_{x \in [0,1]}f(x) - \min_{x \in [0,1]}f(x) &= \max\{f(0),f(1)\} - (1-b^2)\\ &= f(0) - (1-b^2)\\ &= b^2 \end{align}$

This does not equal $4$ when the x-coordinate of the vertex, b, is in the interval $(\frac 12, 0]$.

So the requested sum is -$\frac 32 + \frac 52 = 1$

Steven Alexis Gregory
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