- Samantha has some whiskey at a party, and (being a science and math geek) esti
- Samantha has some whiskey at a party, and (being a science and math geek) estimates her blood alcohol content (BAC) follows the function:
[latex]BAC(t) = \frac{0.25 t}{e^t},[/latex]
where [latex]t[/latex] is measured in hours after her first drink. Graph this function, and determine the following using a derivative:
- How quickly is her BAC increasing (or decreasing) 15 minutes after her first drink?
[latex]BAC'(t) = \frac{0.25 - 0.25t}{e^t}[/latex], [latex]BAC'(1/4) \approx 0.146[/latex] (grams per dL per hour).ans
- How quickly is her BAC increasing (or decreasing) 1 hour after her first drink?
[latex]0[/latex] changeans
- How quickly is her BAC increasing (or decreasing) 2 hours after her first drink?
[latex]\approx 0.034[/latex] grams per dL per hourans
- How quickly is her BAC increasing (or decreasing) 15 minutes after her first drink?
- Graph each function over the given interval. Use calculus to determine the location of all global and local mins and maxes.
- [latex]f(x) = -x^2 + 5x - 2[/latex] on the interval [latex][0, 5][/latex].
Local and Global Mins: [latex](0, -2), (5, -2)[/latex], Local and global max: [latex](2.5, 4.25)[/latex]ans
- [latex]f(x) = x^2 - 6x + 10[/latex] on the interval [latex][2, 6][/latex].
Local min: [latex](2, 2)[/latex], global and local min: [latex](3, 1)[/latex], local and global max: [latex](6, 10)[/latex]ans
- [latex]f(x) = x^3 - 6x^2 + 11x - 6[/latex] on the interval [latex][0, 3][/latex].
Local and global maximum at [latex](1.42, 0.38 )[/latex] , Local min: [latex](2.58,-0.38 )[/latex], local and global minimum: [latex](0, -6)[/latex], local maximum: [latex](3, -2)[/latex]ans
- [latex]f(x) = x^3-5 x^2+8 x-4[/latex] on the interval [latex][0, 2.5][/latex].
Local max at [latex]x = 0.75[/latex], local min at [latex]x = 2[/latex], global max at [latex]x = 2.5[/latex], global min at [latex]x = 0[/latex].ans
- [latex]f(x) = x^4 - 16x[/latex] on the interval [latex][0, 3][/latex].
Local max at [latex]x = 0[/latex], global min at [latex]x = \sqrt[3]{4}[/latex], global max at [latex]x = 3[/latex]ans
- [latex]f(x) = \frac{x^2}{x + 1}[/latex] on the interval [latex][-3, 3][/latex].
Local min at [latex]x = -3[/latex], local max at [latex]x = -2[/latex], local min at [latex]x = 0[/latex], local max at [latex]x = 3[/latex]ans
- [latex]f(x) = -x^2 + 5x - 2[/latex] on the interval [latex][0, 5][/latex].
- Using a chemotherapy drug on a petri-dish of cancer cells, it is found that [latex]P(x)[/latex] percent more of the cancer cells are killed using [latex]x[/latex] milligrams of drug per square centimeter than healthy cells, where [latex]x[/latex] ranges from [latex]0[/latex] to [latex]4[/latex]. It is thought
[latex]P(x) = x^3-8 x^2+16 x[/latex]
For what value of [latex]x[/latex] is [latex]P(x)[/latex] maximized?
[latex]x = 4/3[/latex]ans
- Bananas as we know them may be doomed! Suppose the fungus Tropical Race 4 mentioned in the article is killing off bananas on an island in Jamaica. The number of viable banana farms