1. Diketahui F(x)= [f'(x) dengan f'(x)=3x+4x-3 dan F(2)=-8. Nilai fungsi F(-1) adalah…

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2. Diketahui lim x→3(x²-px+5)=2. Nilai p yang memenuhi adalah…

3. Diketahui lim x→a (x²+6x+4)= 11, nilai a adalah…
NO NGASAL ​

1. Diketahui F(x)= [f'(x) dengan f'(x)=3x+4x-3 dan F(2)=-8. Nilai fungsi F(-1) adalah…

Jawaban:

1.) f'(x) = 3x + 4x – 3

f'(x) = 7x – 3

f(x) = displaystylesfint(7x-3)~dx

f(x) = sffrac{7}{1+1}{x}^{1+1}-3x+C

f(x) = sffrac{7}{2}{x}^{2}-3x+C

f(2) = sffrac{7}{2}{(2)}^{2}-3(2)+C

-8 = sffrac{7}{cancel{2}}cancel{(4)}-6+C

-8 = 7(2) – 3 + C

-8 = 14 – 6 + C

-8 = 8 + C

-8 – 8 = C

C = -16

f(x) = sffrac{7}{2}{x}^{2}-3x-16

f(-1) = sffrac{7}{2}{(-1)}^{2}-3(-1)-16

f(-1) = sffrac{7}{2}(1)-(-3)-16

f(-1) = sffrac{7}{2}+3-16

f(-1) = sffrac{7}{2}-13

f(-1) = sffrac{7}{2}-frac{13times2}{1times2}

f(-1) = sffrac{7}{2}-frac{26}{2}

f(-1) = sffrac{7-26}{2}

f(-1) = sffrac{-19}{2}

f(-1) = sf-9frac{1}{2}

sf~

2.) sflimlimits_{xto3}({x}^{2}-px+5)=2

3² – p3 + 5 = 2

(3×3) – 3p + 5 = 2

9 – 3p + 5 = 2

14 – 3p = 2

-3p = 2 – 14

-3p = -12

p = -12 ÷ (-3)

p = 12 ÷ 3

p = 4

sf~

sflimlimits_{xto{a}}({x}^{2}+6x+4)=11

a² + 6a + 4 = 11

a² + 6a = 11 – 4

a² + 6a = 7

a² + 6a – 7 = 0

a² + 7a – a – 7 = 0

a(a + 7) – (a + 7) = 0

(a + 7)(a – 1) = 0

a + 7 = 0

a – 1 = 0

a = -7

a = 1

asf_{1} = -7 , asf_{2} = 1