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Mathematics

1 answer:

STatiana [176]1 year ago

7 0

To identify the grpah correspondig to the given function find the zeros of the function (use the quadratic formula):

$\begin{gathered} f(x)=\frac{1}{2}x^2+2x-6 \\ \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \\ a=\frac{1}{2} \\ \\ b=2 \\ c=-6 \\ \\ \\ x=\frac{-2\pm\sqrt[]{2^2-4(\frac{1}{2})(-6)}}{2(\frac{1}{2})} \\ \\ x=\frac{-2\pm\sqrt[]{4-(-12)}}{1} \\ \\ x=\frac{-2\pm\sqrt[]{16}}{1} \\ \\ x=\frac{-2\pm4}{1} \\ \\ x_1=-2+4=2 \\ x_2=-2-4=-6 \end{gathered}$

As the zeros are 2 and -6 the graph cross the x-axis in x=2 and x=-6

<h2>Answer: D</h2>

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After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modelled by the function C(t)=8(e

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Answer:

the maximum concentration of the antibiotic during the first 12 hours is 1.185 $\mu g/mL$ at t= 2 hours.

Step-by-step explanation:

We are given the following information:

After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modeled by the function where the time t is measured in hours and C is measured in $\mu g/mL$

$C(t) = 8(e^{(-0.4t)}-e^{(-0.6t)})$

Thus, we are given the time interval [0,12] for t.

We can apply the first derivative test, to know the absolute maximum value because we have a closed interval for t.
The first derivative test focusing on a particular point. If the function switches or changes from increasing to decreasing at the point, then the function will achieve a highest value at that point.

First, we differentiate C(t) with respect to t, to get,

$\frac{d(C(t))}{dt} = 8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)})$

Equating the first derivative to zero, we get,

$\frac{d(C(t))}{dt} = 0\\\\8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0$

Solving, we get,

$8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0\\\displaystyle\frac{e^{-0.4}}{e^{-0.6}} = \frac{0.6}{0.4}\\\\e^{0.2t} = 1.5\\\\t = \frac{ln(1.5)}{0.2}\\\\t \approx 2$