Find a function whose graph has given properties

by RoRi

February 23, 2016

Let $f(x)$ be a function on the interval $[0,1]$ that is nonnegative and differentiable, and such that $f(1) = 0$ . If for each $a$ in the open interval $(0,1)$ the line given by the equation $x = a$ divides the ordinate set of $f(x)$ into regions $A$ and $B$ where $A$ denotes the leftmost region. If the areas of the two regions obey the equation

$A - B = 2f(a) + 3a + b,$

where $b$ is a constant that does not depend on $a$ , find $f(x)$ and $b$ .

Incomplete.

2 comments

March 28, 2019 at 2:25 pm

Evangelos says:

First, let’s define the regions A and B as functions of the variable a

$\begin{align*} A = \int_{0}^{a} f(x) dx \\ B = \int_{a}^{1} f(x) dx \\ \end{align*}$

Now, we take our given equation:

$\begin{align*} A - B = 2f(a) + 3a + b \\ \int_{0}^{a} f(x) dx - \int_{a}^{1} f(x) dx = 2f(a) + 3a + b \\ \int_{0}^{a} f(x) dx + \int_{1}^{a} f(x) dx = 2f(a) + 3a + b \\ \end{align*}$

If we take the derivative with respect to a of both sides of the equation, we get

$\begin{align*} 2f(a) = 2f'(a) + 3 \\ f'(a) = f(a) - \frac{3}{2} \\ \end{align*}$

This equation is satisfied by the function

$\begin{align*} f(a) = Ce^{a} + \frac{3}{2} \\ \end{align*}$

Referring back to our initial condition that f(1) = 0, we can solve for the constant C

$\begin{align*} f(1) = Ce + \frac{3}{2} \\ = 0 \end{align*}$

$\begin{align*} C = -\frac{3}{2e} \end{align*}$

$\begin{align*} f(a) = -\frac{3}{2e}e^{a} + \frac{3}{2} \\ = \frac{3}{2}(1 - e^{a-1}) \end{align*}$

As requested.

Reply
- March 31, 2019 at 1:50 pm
  
  Evangelos says:
  
  And to find the scalar constant b, all we need to do is use our newly found function and evaluate the above integrals.
  
  Recall,
  
  $\begin{align*} A = \int_{0}^{a} f(x) dx \\ B = \int_{a}^{1} f(x) dx \end{align*}$
  
  $\begin{align*} A - B = 2f(a) + 3a + b \end{align*}$
  
  And we now now that the unknown function f(x):
  
  $\begin{align*} f(x) = \frac{3}{2}(1 - e^{x-1}) \end{align*}$
  
  The above equation becomes
  
  $\begin{align*} A - B &= \int_{0}^{a} f(x) dx - \int_{a}^{1} f(x) dx\\ &= \int_{0}^{a} f(x) dx + \int_{1}^{a} f(x) dx \\ &= \frac{3}{2} \int_{0}^{a} (1 - e^{x-1}) dx + \frac{3}{2} \int_{1}^{a} (1 - e^{x-1}) dx \\ &= \frac{3}{2}[x - e^{x-1}]_{0}^{a} + \frac{3}{2}[x - e^{x-1}]_{1}^{a}\\ & = \frac{3}{2}[(a - e^{a-1}) + \frac{1}{e}] + \frac{3}{2}[(a - e^{a-1})]\\ & = 3(a - e^{a-1}) + \frac{3}{2e} \\ &= 2f(a) + 3a + b \end{align*}$
  
  $\begin{align*} 3(a - e^{a-1}) + \frac{3}{2e} &= 3(1 - e^{a-1}) + 3a + b\\ \end{align*}$
  
  $\begin{align*} \frac{3}{2e} &= 3 + b\\ \end{align*}$
  
  $\begin{align*} b = \frac{3}{2e} - 3 \quad \blacksquare \end{align*}$
  
  Reply