f(x) = 3/2x + 1, with x= a, and x = a + h?

Method 1: substitute x+h for x.

a)

f(x+h)=2(x+h)+3=2x+2h+3

f(x)=2x+3

[f(x+h)-f(x)]/h=(2x+2h+3-2x-3)/h=2h/h=...

b)

f(x+h)=1/[x+h+1]

f(x)=1/(x+1)

So, we have:

1/[a+b]-1/[a]=

[a]/[a^2+ab]-[a+b]/[a^2+ab]=

[a-a-b]/[a^2+ab]=

[-b]/[a^2+ab]

for a=x+1, b=h,

[-b]/[a^2+ab]=

-h/[x^2+2x+1+xh+h]

then we have:

(a/b)/c=a/(bc)

for a=-c,

-c/bc=(-c/c)(1/b)=

-1/b

for a=-h, c=h, b=x^2+2x+1+xh+h

now the limit of h approaching 0 causes this to become:

b=x^2+2x+1=(x+1)^2

-1/(x+1)^2=

-(x+1)^-2

c)

f(x+h)=(x+h)^2=x^2+2xh+h^2

f(x+h)-f(x)=x^2+2xh+h^2-x^2=2xh+h^2

Divide this by h to get:

2x+h

limit as h approaches 0 makes this:

2x

Method 2: just find the derivative

f(x)=x^a

the derivate:

f'(x)=ax^(a-1)

and if there are more terms we find the derivative of each separate term.

So,

a)

f(x)=2x+3

f(x)=2x^1+3x^0

f'(x)=(2*1)x^(1-1)+(3*0)x^(0-1)

f'(x)=2x^0+0

f'(x)=2

b)

f(x)=1/(x+1)

f(x)=(x+1)^-1

f(x)=1(x+1)^-1

f'(x)=(1*-1)(x+1)^(-1-1)

f'(x)=-1(x+1)^-2

f'(x)=-(x+1)^-2

c)

f(x)=x^2

f(x)=1x^2

f'(x)=(1*2)x^(2-1)

f'(x)=2x^1

f'(x)=2x

May be

Bieber is actual in his calculation of the inverses of the two applications, and the area and selection of the 1st. to locate the area of the 2nd function, photograph (or comedian strip) the graph of e^x. As x gets further and extra unfavorable, it procedures the x axis from the effective y area. It crosses the y-axis at (0,a million), then is going upwards. The area of this function is all actual numbers, and the type is y > 0. So, on your problem, the area is all actual numbers and the type is y > 0.