What is the local linearization of f(x) = e^x^5 near x = 1?

Question

I approached this problem by assuming local linearization is the same as finding the tangent. I plugged 1 into the function and got e, so I know later in my formula to use the point (1,e)

Then I took the derivative and plugged in one and got 5e. This I used for a slope, and I got the equation

5ex-5e+e, so I then reduced to get 5ex-4e. But I got marked wrong on this, and I don't know how to solve this any other way. Please help me.

Answer 1

The formula for local linearization is

 

L(x) = f(x_o) + f'(x₀)(x - x₀)

 

where  x₀ = 1

 

 

Find the derivative of f(x).

 

Let  r = x⁵

 

then  f(x) = e^r

 

f'(x) = r'e^r

 

f'(x) = 5x⁴(e^x⁵)

 

f'(1) = 5(1)⁴(e^(1⁵))

 

f'(1) = 5e

 

 

Then f(1) = e

 

 

Therefore,

 

L(x) = e + 5e(x - 1)