I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
Section 3.4 : Product and Quotient Rule
9. Find the equation of the tangent line to \(f\left( x \right) = \left( {1 + 12\sqrt x } \right)\left( {4 - {x^2}} \right)\) at \(x = 9\).
Show All Steps Hide All Steps
Start SolutionWe know that the derivative of the function will give us the slope of the tangent line so we’ll need the derivative of the function. We’ll use the product rule to get the derivative.
\[f'\left( x \right) = \left( {6{x^{ - \,\,\frac{1}{2}}}} \right)\left( {4 - {x^2}} \right) + \left( {1 + 12\sqrt x } \right)\left( { - 2x} \right) = \left( {\frac{6}{{\sqrt x }}} \right)\left( {4 - {x^2}} \right) - 2x\left( {1 + 12\sqrt x } \right)\] Show Step 2Note that we didn’t bother to “simplify” the derivative (other than converting the fractional exponent back to a root) because all we really need this for is a quick evaluation.
Speaking of which here are the evaluations that we’ll need for this problem.
\[f\left( 9 \right) = \left( {37} \right)\left( { - 77} \right) = - 2849\hspace{0.5in}f'\left( 9 \right) = \left( 2 \right)\left( { - 77} \right) - 18\left( {37} \right) = - 820\] Show Step 3Now all that we need to do is write down the equation of the tangent line.
\[y = f\left( 9 \right) + f'\left( 9 \right)\left( {x - 9} \right) = - 2849 - 820\left( {x - 9} \right)\hspace{0.25in} \to \hspace{0.25in}\,\require{bbox} \bbox[2pt,border:1px solid black]{{y = - 820x + 4531}}\]