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Assuming you mean
first use chain rule on first part
derivitive of (x+1)⁵=5(x+1)⁴ times the derivitive of (x+1) which is 1
the first part is 5(x+1)⁴
derivitive of -5x is -5
derivitive of -2 is 0
we gots
f'(x)=5(x+1)⁴-5
find zeroes
0=5(x+1)⁴-5
5=5(x+1)⁴
1=(x+1)⁴
+/-1=x+1
-1+/-1=x
-2=x or 0=x
f'(x)=5(x+1)⁴-5
zeroes at x=-2 and x=0
first use chain rule on first part
derivitive of (x+1)⁵=5(x+1)⁴ times the derivitive of (x+1) which is 1
the first part is 5(x+1)⁴
derivitive of -5x is -5
derivitive of -2 is 0
we gots
f'(x)=5(x+1)⁴-5
find zeroes
0=5(x+1)⁴-5
5=5(x+1)⁴
1=(x+1)⁴
+/-1=x+1
-1+/-1=x
-2=x or 0=x
f'(x)=5(x+1)⁴-5
zeroes at x=-2 and x=0
Answer:
Step-by-step explanation:
a) 4x² + 20 = 0
4x² = -20
x² = -20/4
x² = -5 = i² *5
x = √i²5
x = i√5
b) 4x² + 5 = -7
4x² = -7 - 5
4x² = - 12
x² = -12/4
x² = -3 = 3i²
x = i√3
c) x² + 4x + 20 = 0
a = 1 ; b =4 ; c =20
= (-2 +4i) & -12/2
= (-2 + 4i) and (-6)