What is the equation of the line tangent to # f(x)=1/sqrt(e^x-3x) # at # x=0#?

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Answer 1 · 2017-08-18T22:48:20

The equation is #y = x + 1#.

At #x= 0#, the function has y-value

#f(0) = 1/sqrt(e^0 - 3(0)) = 1/sqrt(1) = 1#

We now rewrite the function as

#f(x) = (e^x- 3x)^(-1/2)#

We can find this derivative using the chain rule.

#f'(x) = -1/2u^(-3/2) * (e^x - 3) = -(e^x - 3)/(2(e^x - 3x)^(3/2)#

The slope of the tangent is hence

#f'(0) = -(e^0 - 3)/(2(e^0 - 3(0))^(3/2))#

#f'(0) = - (-2)/(2)#

#f'(0) = 2/2#

#f'(0) = 1#

Now write the equation of the tangent.

#y -y_1 = m(x - x_1)#

#y - 1 = 1(x - 0)#

#y = x + 1#

Hopefully this helps!