#### To determine

**(a)**

**To find:**f'(x), graph f .

#### Answer

f'x=-3x+1025-x

#### Explanation

**1) Formula:**

i. Difference rule:

ddxfx-gx=ddxfx-ddxg(x)

ii. Power rule combined with chain rule:

ddxf(x)n=nfxn-1.f'(x)

iii. constant function rule:

ddxc=0 where c is constant.

**2) Formula:**

Equation of tangent line to the curve y=f(x) at point (a,f(a)) is,

y-fa=f’ax-a

**3) Given:**

y=fx=x5-x

**4) Calculation:**

Differentiate given equation with respect to x,

f'x=ddxx5-x

Using product rule,

ddxx5-x=xddx5-x+5-xddxx

By using power rule combined with chain rule,

f'x=x25-x.ddx(5-x)+5-x

Using difference rule, constant function rule, power rule

f'x=-x25-x+5-x=-3x+1025-x

**Final statement:**

f'x=-3x+1025-x

#### To determine

**(b)**

**To find: **equation of tangent line.

#### Answer

Equation of tangent line at (1,2) is

y=74x+14

Equation of tangent line at (4,4) is y=-x+8

#### Explanation

**1) Calculation:**

At x=1

f’1=-3x+1025-x=74

So the equation of tangent line to the curve y=x5-x at point (1,2) is,

y-2=f’1x-1=74x-1=7x4-74

Adding 2 to both sides,

y=7x4-74+2

y=7x4+14

At x=4

f’4=-3x+1025-x=-22=-1

Equation of tangent line to the curve y=x5-x at point (4,4) is,

y-4=f’4x-4=-1x-4=-x+4

Adding 4 to both sides,

y=-x+8

**Final statement:**

Equation of tangent line at (1,2) is

y=74x+14

Equation of tangent line at (4,4) is

y=-x+8

#### To determine

**(c)**

**To illustrate:** Part(b) by graphing curve and tangent lines

#### Explanation

#### To determine

**(d)**

**To check:** answer to part(a) is reasonable by comparing the graphs f and f’

#### Explanation

In the figure green is graph of f(x) and blue is graph of f’(x). Notice that when f is increasing f’ is positive and when f is decreasing f’(x) is negative. Also when f has a horizontal tangent f’ is zero. So the f’ we computed has all properties of derivative of f. So it is a reasonable answer.

Therefore, answer to part(a) is reasonable by comparing the graphs f and f’