cal 2 help - min value and "simple" derivative?

Question

i need input on these 2 problems. i've been on them for over 45 mins and can't solve them.

1. give mininum value of the function:

f(x) = 7e^(5x) + 5e^(-5x)

2. what is the derivative of f at x=0?

f(x) = 5xe^(4x + 2ln2)

i keep getting zero for no. 2. i know what's not the correct answer. =(

TIA!!!

Answer 1

1. f(x) = 7e^(5x) + 5e^(-5x)
df/dx = 35e^(5x) - 25e^(-5x) = 0

Multiply through by e^(5x):

35e^(5x + 5x) - 25 = 0

e^(10x) = 25/35 = 5/7

10x = ln(5/7)

x = ln(5/7) / 10

Now we need to evaluate f(x) at that point.

First, e^(5x) = e^[ln(5/7) / 2] = sqrt(5/7)
and e^(-5x) = e^[-ln(5/7) / 2] = 1/ sqrt(5/7) = sqrt(7/5)

f(min) = 7 sqrt(5/7) + 5 sqrt(7/5)
f(min) = sqrt(35) + sqrt(35) = 2 sqrt(35) (Answer)


2. f(x) = 5xe^(4x + 2 ln 2) = 5x e^(4x) e^(2 ln 2)
f(x) = 5x e^(4x) e^(ln 4)
f(x) = 4*5x e^(4x) = 20x e^(4x)
df/dx = 20x[4 e^(4x)] + e^(4x) (20)
df/dx = e^(4x) (80x + 20)

For x=0: df/dx = 20 (Answer)

Answer 2

2. f'=20xe^(4x+2ln2)+5*e^(4x+2ln20)
f'(0)= 5*e^2ln20 just work that out

1. f'=35e^(5x)-25e^(-5x), set =0

7e^(5x)=5e^(-5x)
e^(10x)=5/7 10x=ln5-ln7 solve for x.
check some other values to see max or min