Question

and draw the tangent lines to the graph at points whose x-coordinates are -2 , 0, and 1.

Answer 1

a)

To find the difference quotient we will do

$\frac{f(x+h)-f(x)}{h}$

But f(x) = 4x, therefore f(x+h) = 4(x+h). Then

$\frac{f(x+h)-f(x)}{h}=\frac{4(x+h)-4x}{h}$

Now we just simplify the expression

$\frac{4(x+h)-4x}{h}=\frac{4x+4h-4x}{h}=\frac{4h}{h}=4$

Therefore

$\frac{f(x+h)-f(x)}{h}=4$

As expected, because the function f is a linear function, then it must be a constant.

b)

As we can see we don't have the term "h" in the difference quotient, then we can easily solve the limit by just repeating the result:

$\lim _{h\rightarrow0}\frac{f(x+h)-f(x)}{h}=\lim _{h\rightarrow0}4=4$

Therefore

$f^{\prime}(x)=4$

See that it's a constant function, it means that the slope will always be the same, doesn't matter the value of x.