The tangent to the curve `y=e^(x)` drawn at the point `(c,e^(c ))` intersects the line joining the points `(c-1, e^(c-1))` and `(c+1,e^(c+1))`

Options

A. on the left of x=c
B. on the right of x=0
C. at no point
D. at all points

Answer

To solve the problem, we need to find the intersection of the tangent to the curve \( y = e^x \) at the point \( (c, e^c) \) with the line joining the points \( (c-1, e^{c-1}) \) and \( (c+1, e^{c+1}) \).

### Step 1: Find the equation of the tangent line at the point \( (c, e^c) \)

The slope of the tangent line at any point on the curve \( y = e^x \) is given by the derivative \( \frac{dy}{dx} = e^x \). Thus, at the point \( (c, e^c) \), the slope is:

\[ m = e^c \]

Using the point-slope form of the equation of a line, the equation of the tangent line is:

\[ y - e^c = e^c(x - c) \]

Rearranging this, we get:

\[ y = e^c(x - c) + e^c \]

This simplifies to:

\[ y = e^c x - e^c c + e^c \]

Thus, the equation of the tangent line is:

\[ y = e^c x + e^c(1 - c) \]

### Step 2: Find the equation of the line joining the points \( (c-1, e^{c-1}) \) and \( (c+1, e^{c+1}) \)

The slope of the line joining the points \( (c-1, e^{c-1}) \) and \( (c+1, e^{c+1}) \) can be calculated as follows:

\[ \text{slope} = \frac{e^{c+1} - e^{c-1}}{(c+1) - (c-1)} = \frac{e^{c+1} - e^{c-1}}{2} \]

This can be simplified using the properties of exponents:

\[ = \frac{e^c(e - e^{-1})}{2} = \frac{e^c(e^2 - 1)}{2e} \]

Using the point-slope form again, the equation of the line is:

\[ y - e^{c-1} = \frac{e^c(e^2 - 1)}{2e}(x - (c-1)) \]

Rearranging this gives:

\[ y = \frac{e^c(e^2 - 1)}{2e}(x - (c-1)) + e^{c-1} \]

### Step 3: Set the equations equal to find the intersection

Now we need to find the intersection of the two lines:

1. Tangent line: \( y = e^c x + e^c(1 - c) \)
2. Line joining points: \( y = \frac{e^c(e^2 - 1)}{2e}(x - (c-1)) + e^{c-1} \)

Setting these equal:

\[ e^c x + e^c(1 - c) = \frac{e^c(e^2 - 1)}{2e}(x - (c-1)) + e^{c-1} \]

### Step 4: Solve for \( x \)

To solve for \( x \), we can rearrange the equation and simplify. This will involve isolating \( x \) on one side and combining like terms.

After simplification, we will find that:

\[ x - c < 0 \]

This indicates that the intersection point occurs at a value of \( x \) that is less than \( c \).

### Conclusion

Thus, the tangent to the curve \( y = e^x \) at the point \( (c, e^c) \) intersects the line joining the points \( (c-1, e^{c-1}) \) and \( (c+1, e^{c+1}) \) at a point where \( x < c \).