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#

The given function is #f(x) = e^(x)#.

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

Let's find this #y#-intercept.

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

#Rightarrow f(0) = e^(0) = 1#

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

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

Let's differentiate #f(x)#.

The derivative of #e^(x)# is a standard derivative: it's #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#.