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

Question

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 $4$.

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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.

Answer 1

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 $-\frac 32+\frac 52=\color{red}{1}$.

Answer 2

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 -$\frac 32 + \frac 52 = 1$