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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
First off, let's notice that the squared variable is the "x", that means is a vertically opening parabola.
notice the leading term's coefficient, is -2, is negative, that means the parabola is opening downwards, is facing down, so it goes up up up, reaches the U-turn and then down down down, is a "hump" or a maximum point.
and the line of symmetry, well is a vertical parabola, mirroring itself at the line x = -2, which is the x-coordinate of the vertex.
notice the leading term's coefficient, is -2, is negative, that means the parabola is opening downwards, is facing down, so it goes up up up, reaches the U-turn and then down down down, is a "hump" or a maximum point.
and the line of symmetry, well is a vertical parabola, mirroring itself at the line x = -2, which is the x-coordinate of the vertex.