SOLUTION: f(x) =-3e^x+3 + e^1 find f'(0)

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Question 1180614: f(x) =-3e^x+3 + e^1 find f'(0)
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

f%28x%29=+-3e%5E%28x%2B3%29+%2B+e%5E1
Find f'%280%29
first find f'%28x%29

%28d%2Fdx%29%28-3e%5E%28x%2B3%29+%2B+e%5E1%29.......apply the Sum/Difference Rule
=%28d%2Fdx%29%28-3e%5E%28x%2B3%29+%29+%2B+%28d%2Fdx%29%28e%5E1%29
=-3e%5E%28x%2B3%29%2B+0

=-3e%5E%28x%2B3%29

f'%28x%29+=+-3e%5E%28x%2B3%29
then
f'%280%29+=+-3e%5E%280%2B3%29
f'%280%29+=+-3e%5E3->exact
f'%280%29+=+-60.26->approximately

Answer by ikleyn(52800) About Me  (Show Source):
You can put this solution on YOUR website!
.

Hello, are you sure that the last addend in the formula is   e^1 ?


I am asking because  NOBODY  and  NEVER  uses such a form of writing in  Math . . .