Question

Find the linearization L(x) of y=e8xln(x) at a=1

Answer 1

Answer: $L(x) = e^8 (x-1)$

Explanation:

The linearization L(x) of y = f(x) at x = a is given by

$L(x) = f(a) + f'(a) (x - a)$     (1)

Where f'(a) is the derivative of y = f(x) evaluated at x = a. So, we need to find the derivative of y = f(x) and evaluate the derivative at x = a. 

The derivative of y = f(x) is computed using the product rule for the derivatives because f(x) is the product of two functions: logarithmic and exponential. 

So, the derivative of y = f(x) is given by

$f'(x) = (\frac{d}{dx} (e^{8x}))(\ln x) + (\frac{d}{dx} (\ln x))(e^{8x}) \\ \\ f'(x) = 8e^{8x}\ln x + (e^{8x})(\frac{1}{x}) \\ \\\boxed {f'(x) = 8e^{8x}\ln x + \frac{e^{8x}}{x}}$

Since a = 1, the derivative of y = f(x) evaluated at x = a = 1 is given by

$f'(a) = f'(1) \\ f'(a) = 8e^{8(1)}\ln 1 + \frac{e^{8(1)}}{1} \\ \\ f'(a) = 8e^{8}(0) + e^{8} \\ \boxed{f'(a) = e^8}$

Moreover, note that

$f(a) = f(1) \\ f(a) = e^{8(1)}(\ln 1) \\ f(a) = e^8 (0) \\ \boxed{f(a) = 0}$

Using equation (1), the linearization L(x) of y = f(x) at x = a = 1 is given by

$L(x) = f(a) + f'(a) (x - a) \\ L(x) = 0 + e^8 (x - 1) \\ \boxed{L(x) = e^8 (x - 1)}$