Animated Solution for Mathematics - Differentiation: 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 $(c+1,e^{c+1})$

Welcome, fellow traveler on the path to JEE mastery. Today, we are exploring the elegant nature of the exponential function $y=e^x$. This curve is the heartbeat of calculus, and understanding its geometry is a vital skill.

Imagine standing on the coordinate plane, looking at the curve $y=e^x$. It is a beautiful, ever-rising slope, and we are going to dissect the relationship between a tangent and a secant line.

Our journey begins at point $P(c,e^c)$. To find the tangent line, we need its slope. As you know, the derivative of $f(x)=e^x$ is simply $f^{\prime }(x)=e^x$.

At $x=c$, the slope $m_T$ is $e^c$. Using the point-slope form, $y-y_1=m(x-x_1)$, we get the equation of our tangent line:

Now, let us look at the secant line. It connects two points on the curve: $A(c-1,e^{c-1})$ and $B(c+1,e^{c+1})$. The slope of this secant, $m_S$, is the change in $y$ over the change in $x$:

Using point $A$, the equation of the secant line becomes:

To find where these two lines meet, we equate their $y$-values. Let $t=x-c$. This substitution is our secret weapon.

The tangent equation becomes $y=e^c(t+1)$. The secant equation becomes:

Equating them and dividing by $e^c$ yields:

Solving for $t$, we find:

Now, look at the numerator: $e+e^{-1}-2$. This is $(\sqrt{e}-\frac{1}{\sqrt{e}}{)}^2$, which is strictly positive.

The denominator, $2-e+e^{-1}$, is approximately $2-2.718+0.368$, which is negative. A positive divided by a negative is negative.

Thus, $t=x-c<0$, which means $x<c$. The intersection point lies to the left of $x=c$. You have just conquered the geometry of the exponential curve.