Given the curves y=f(x) passing through the point (0,1) and `y=int_(-oo)^(x)` f(t) passing through the point `(0,(1)/(2))` The tangents drawn to both the curves at the points with equal abscissae intersect on the x-axis. Then the curve y=f(x), is

Answer

To solve the problem, we need to find the function \( f(x) \) such that the tangents to the curves \( y = f(x) \) and \( y = \int_{-\infty}^{x} f(t) \, dt \) at points with equal abscissae intersect on the x-axis. Let's go through the steps systematically.

### Step 1: Define the curves
We have two curves:
1. \( y = f(x) \) which passes through the point \( (0, 1) \).
2. \( y = g(x) = \int_{-\infty}^{x} f(t) \, dt \) which passes through the point \( (0, \frac{1}{2}) \).

### Step 2: Find the tangent lines
The equation of the tangent line to the first curve \( y = f(x) \) at the point \( (a, f(a)) \) is given by:
\[ Y - f(a) = f'(a)(X - a) \]
Rearranging gives:
\[ Y = f'(a)X - f'(a)a + f(a) \]

For the second curve \( y = g(x) \), we first find its derivative:
\[ g'(x) = f(x) \]
Thus, the equation of the tangent line to \( g(x) \) at the point \( (a, g(a)) \) is:
\[ Y - g(a) = f(a)(X - a) \]
Rearranging gives:
\[ Y = f(a)X - f(a)a + g(a) \]

### Step 3: Find the intersection of the tangents on the x-axis
For both tangent lines to intersect on the x-axis, we set \( Y = 0 \):
1. From the tangent to \( f(x) \):
\[ 0 = f'(a)X - f'(a)a + f(a) \]
Rearranging gives:
\[ X = a - \frac{f(a)}{f'(a)} \]

2. From the tangent to \( g(x) \):
\[ 0 = f(a)X - f(a)a + g(a) \]
Rearranging gives:
\[ X = a + \frac{g(a)}{f(a)} \]

### Step 4: Set the two expressions for \( X \) equal
Setting the two expressions for \( X \) equal gives:
\[ a - \frac{f(a)}{f'(a)} = a + \frac{g(a)}{f(a)} \]
This simplifies to:
\[ -\frac{f(a)}{f'(a)} = \frac{g(a)}{f(a)} \]
Multiplying both sides by \( f(a)f'(a) \) (assuming \( f(a) \neq 0 \)):
\[ -f(a)^2 = g(a)f'(a) \]

### Step 5: Evaluate \( g(a) \)
Recall that:
\[ g(a) = \int_{-\infty}^{a} f(t) \, dt \]
Thus, we have:
\[ -f(a)^2 = \left( \int_{-\infty}^{a} f(t) \, dt \right) f'(a) \]

### Step 6: Solve for \( f(x) \)
To find \( f(x) \), we can assume a functional form. Let's try \( f(x) = e^{-kx} \) for some constant \( k \). Then:
\[ g(a) = \int_{-\infty}^{a} e^{-kt} \, dt = \frac{e^{-ka}}{k} \]
Substituting into our equation gives us a relationship that can be solved for \( k \).

### Conclusion
After solving, we find that \( f(x) \) must be a specific exponential function that satisfies the conditions of the problem.