Find an equation for the line tangent to the graph of
f(x)=−4xe^x
at the point (a,f(a)) for a=1.
y=
I'm unable to solve this problem I tried so many times but I didn't get the answer?
2017-01-17 9:23 am
回答 (4)
2017-01-17 9:23 am
✔ 最佳答案
Is 15 
2017-01-17 9:28 am
f(x) = −4x e^x
f(1) = −4e
f'(x) = −4e^x − 4xe^x = −4e^x(x+1)
f'(1) = −8e
Tangent line passes through point (1,−4e) and has slope = −8e
y + 4e = −8e(x − 1)
y = −8ex + 8e − 4e
y = −8ex + 4e
2017-01-17 9:26 am
hint:
tangent
y-f(a)=f'(a)(x-a)
--->answer
y=-8ex+4e
tangent
y-f(a)=f'(a)(x-a)
--->answer
y=-8ex+4e
收錄日期: 2021-04-20 17:59:13
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