# 2 calculus problems

• Mar 3rd 2008, 06:39 AM
salu34
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!
• Mar 3rd 2008, 07:02 AM
colby2152
Quote:

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):

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

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

\$\displaystyle f'(1)=6\$

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

\$\displaystyle f(1)=1\$

Use the slope-intercept form for a line:

\$\displaystyle y=mx+b\$

Plug and chug...

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

\$\displaystyle y=6x-5\$
• Mar 3rd 2008, 07:28 AM
salu34
?
how did you find that the derivative of 6(2x-1)^2 was 6?
• Mar 3rd 2008, 07:48 AM
Simplicity
Quote:

Originally Posted by salu34
how did you find that the derivative of 6(2x-1)^2 was 6?

colby2152 substituted \$\displaystyle x=1\$ into the differentiated function.

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