Question

Turunan Fungsi
diketahui f(x) = x²+3/2x+1, x≠-1/2. Hasil dari f(0)+f'(0) = . . .
a. -10
b. -9
c. -7
d. -5
e. -3

Answer 1

Turunan

f=(x^2+3)/(2x+1)

u=x^2+3. u'=2x
v=2x+1. v'=2

f'(x)=u'(x).v(x)-u(x).v'(x)
-----------------------
{v(x)}^2
={2x.(2x+1)}-{x^2+3}.(2)}
-----------------------------
(2x+1)^2

={4x+2x}-{2x^2+6}
-----------------------
(4x^2+4x+1)

={6x}-{2x^2+6}
------------------
4x^2+4x+1

=-2x^2+6x-6
---------------
{4x^2+4x+1}

f(0)+f'(0)=?

f(0)=((0)^2+3)/(2(0)+1)
=(0+3)/(0+1)
=3/1
=3
f'(0)=-2(0)^2+6(0)-6
-------------------
4(0)^2+4(0)+1
=0+0-6
---------
0+0+1
=-6/1
=-6

f(0)+f'(0)=3+(-6)=-3
Pilihan E.

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