# Math Help - 2 calculus problems

1. ## 2 calculus problems

1) Find the equation of the tangent to f(x)=(2x-1)^3 where x=1

2) Suppose that for a compant manufacturing calculators, the cost, revenue, and profit equations are given by C=90,000+30x R=300x - (x^2)/30 & P=R-C where the production output in 1 week is x calculators. if production is increasing at a rate of 500 calculators per week, when production output is 6,000 calculators, find the rate of increase (decrease) in: a) cost b) revenue, c) profit.

Any help would be greatly appreciated.
Thanks!

2. Originally Posted by salu34
1) Find the equation of the tangent to f(x)=(2x-1)^3 where x=1

2) Suppose that for a compant manufacturing calculators, the cost, revenue, and profit equations are given by C=90,000+30x R=300x - (x^2)/30 & P=R-C where the production output in 1 week is x calculators. if production is increasing at a rate of 500 calculators per week, when production output is 6,000 calculators, find the rate of increase (decrease) in: a) cost b) revenue, c) profit.

Any help would be greatly appreciated.
Thanks!

Find the derivative of f(x):

$f'(x)=6(2x-1)^2$

The slope is the rate of change at a given point, (i.e. it's derivative).

$f'(1)=6$

Now, what is the y-coordinate for x=1?

$f(1)=1$

Use the slope-intercept form for a line:

$y=mx+b$

Plug and chug...

$1=6(1)+b$
$b=-5$

$y=6x-5$

3. ## ?

how did you find that the derivative of 6(2x-1)^2 was 6?

4. Originally Posted by salu34
how did you find that the derivative of 6(2x-1)^2 was 6?
colby2152 substituted $x=1$ into the differentiated function.

$f'(1) = 6(2x-1)^2$
$f'(1) = 6(2(1)-1)^2$
$f'(1) = 6(2-1)^2$
$f'(1) = 6(1)^2$
$f'(1) = 6$