Find the linearization L(x) of y= e^4x ln(x) at a=1 // L(x) = ...?

Question

Find the linearization L(x) of y= e^4x ln(x) at a=1

L(x) = ...?

Answer 1

Let f(x) = e^4x ln(x) (x> 0)

f(a) = f(1) = 0

That means the linearization for f at the point where x =1 is the equation of the tangent line to f at (1,0).

f'(x) = e^4x/x + ln(x) e^4x(4)

f'(1) = e⁴/1 + e⁴/4 (ln(1)) = e⁴ (since ln(1) = 0)

The slope of the tangent line at the point where x = 1, is e⁴.

The equation of a line with slope e⁴ passing through (1, 0) is:

y = e⁴(x - 1)

or

L(x) = e⁴(x - 1)

At points on f with an x value close to 1, L(x) will give a close approximation for the function value:

f(1.001) ≈ 0.0547895897701

L(1.001) ≈ 0.0545981500331

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