Home > Calculus > Tangent Line to a Curve

What is the equation of the tangent line of #f(x)=ln(x^2+4x+e) # at #x=0#?

Answer 1

#y=(4x)/e+1#

Initially, locate the point where the tangent line will cross:

#f(0)=ln(0^2+4(0)+e)=ln(e)=1#

Hence the point of tangency is #(0,1)#.

All we need to know now is the slope of the tangent line at #x=0#. To do this, find the value of the derivative at #x=0#.

To find the derivative of #f(x)#, we will use the chain rule.

In particular, since

#d/dxln(x)=1/x#

We are aware of that

#d/dxln(g(x))=1/g(x)*g'(x)=(g'(x))/g(x)#

Applying this to #f(x)=ln(x^2+4x+e)#, we see that

#f'(x)=(d/dx(x^2+4x+e))/(x^2+4x+e)=(2x+4)/(x^2+4x+e)#

The tangent line's slope is

#f'(0)=(2(0)+4)/(0^2+4(0)+e)=4/e#

The equation of a line that passes through the point #(0,1)# with a slope of #"4/"e# is

#y=(4x)/e+1#

By plotting the original function and its tangent line, we can verify:

graph{(y-(4x)/e-1)=0 [-.8, 1, -2.21, 4.03]} ln(x^2+4x+e)-y)

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Ezekiel Alexander

The equation of the tangent line of f(x)=ln(x^2+4x+e) at x=0 is y = 1.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.