Hi. I don't quite get this: At the point where the function f(x) = e^x intersects the y-axis the tangent is drawn. Find the x-coordinate of the intercept of this tangent with the x-axis. ?

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Answer 1 · 2018-03-24T13:23:12

$x = - 1$ $x = - 1$

The given function is $f (x) = e^{x}$ $f(x) = e^(x)$ .

A tangent is said to be drawn at the point where $f (x)$ $f(x)$ intersects the $y$ $y$ -axis.

Let's find this $y$ $y$ -intercept.

The $y$ $y$ -intercept occurs when $x = 0$ $x = 0$ :

$\Rightarrow f (0) = e^{0} = 1$ $Rightarrow f(0) = e^(0) = 1$

So, a tangent is drawn at the point where $x = 0$ $x = 0$ and $y = 1$ $y = 1$ , i.e. $(0, 1)$ $(0, 1)$ .

Now, we need to find an equation for the tangent drawn at this point.

Let's differentiate $f (x)$ $f(x)$ .

The derivative of $e^{x}$ $e^(x)$ is a standard derivative: it's $e^{x}$ $e^(x)$ as well:

$Rightarrow f'(x) = e^(x)$

Now, we need to find the gradient of the tangent at $(0, 1)$ .

Let's substitute $0$ in place of $x$ in the derivative:

$Rightarrow f'(0) = e^(0) = 1$

So, the gradient of the tangent is $1$ .

We can now write an equation for the tangent drawn at $(0, 1)$ :

$Rightarrow y - 1 = 1 (x - 0) = x - 0 = x$

$Rightarrow y = x + 1$

Finally, we are required to find the $x$ -coordinate of the intercept of the tangent with the $x$ -axis.

So let's the equation of the tangent equal to zero:

$Rightarrow y = 0$

$Rightarrow x + 1 = 0$

$therefore x = - 1$

Therefore, the tangent intersects the $x$ -axis at $x = - 1$ .