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How do you find the linearization at a=0 of #f(x) = e^(5 x)#?

Answer 1

Emily Ammerman

Linearization uses the Linear Approximation Method.

#f_T(a) = f(a) + f'(a)(x-a)#

So, let's take the derivative first, since that's one thing we don't have. If we let #u = 5x#...

#f'(x) = d/(dx)[e^(u)]*(du)/(dx) = e^(5x) * 5 = 5e^(5x)#

Finally, let's plug in #a = 0#:

#f_T(0) = f(0) + f'(0)*(x)#

#= e^0 + 5*e^0*(x)#

#= 5x + 1#

So the tangent line is just #color(blue)(f_T(0) = 5x + 1)#.

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Answer 2

Axel Alvarez

To find the linearization of ( f(x) = e^{5x} ) at ( a = 0 ), we first find the first derivative of ( f(x) ) with respect to ( x ), evaluate it at ( x = 0 ), and then use that value to form the equation of the tangent line at ( x = 0 ).

The derivative of ( f(x) = e^{5x} ) with respect to ( x ) is ( f'(x) = 5e^{5x} ).

Evaluate ( f'(x) ) at ( x = 0 ): ( f'(0) = 5e^{5(0)} = 5 ).

So, the slope of the tangent line at ( x = 0 ) is ( m = 5 ).

To find the y-intercept of the tangent line, we substitute ( x = 0 ) into ( f(x) ): ( f(0) = e^{5(0)} = 1 ).

Therefore, the equation of the tangent line at ( x = 0 ) is ( y = 5x + 1 ).

This is the linearization of ( f(x) = e^{5x} ) at ( a = 0 ).

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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.