# Determine the differentiable function $f$ whose graph lies above the x-axis and passes through the point $(0, 1)$

I'm trying to solve the question:

Determine the differentiable function $$f$$ whose graph lies above the $$x$$-axis and passes through the point $$(0, 1)$$ and such that for any $$x\geq0$$ the area under the graph of $$y = f(x)$$ above the interval $$[0, x]$$ is four times the slope of the tangent line to the graph of $$y = f(x)$$ at the point $$(x, f(x))$$.

I have written $$\int_0^xf(x)dx=4f'(x)$$ to start with, but I'm not sure if this is correct.

Any help would be appreciated. Thanks in advance.

• That is correct so far, but it is better written as $\int_0^x f(t)dt=4f'(x)$. Commented Sep 19, 2021 at 12:12

## 2 Answers

Clearly since $$\int_0^xf(t)dt=4f'(x)$$ for all $$x\in\mathbb{R}$$, we have that $$f'$$ is differentiable and, $$\frac d{dx}\int_0^xf(t)dt=f(x)=4f''(x)$$

Note that,

The general solution of the differential equation $$4y''-y=0$$ is $$y=A\sinh(\frac x2)+B\cosh(\frac x2)$$ for abitrary $$A,B\in\mathbb{R}$$.

Hence $$f(x)=A\sinh(\frac x2)+B\cosh(\frac x2)$$ for all $$x\in\mathbb{R}$$ for some $$A,B\in\mathbb{R}$$ and since $$f(0)=1$$ we have that $$B=1$$. Also $$\int_0^0f(x)dx=4f'(0)$$gives $$f'(0)=\frac12A=0$$. Therefore the only solution to the equation is $$f:\mathbb{R}\to\mathbb{R}$$ defined by, $$f(x)=\cosh\left(\frac x2\right)\text{ for all }x\in\mathbb{R}$$

Since$$\int_0^xf(t)\,\mathrm dt=4f'(x),\tag1$$differentiating both sides you get that $$f(x)=4f''(x)$$. So, $$f(x)=a\cosh\left(\frac x2\right)+b\sinh\left(\frac x2\right)$$ for some constants $$a$$ and $$b$$. Since $$f(0)=1$$, $$a=1$$. On the other hand, it follows from $$(1)$$ that $$f'(0)=0$$. So, $$b=0$$, and $$f(x)=\cosh\left(\frac x2\right)$$.