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3 years ago

If f(x) = x^2-6x+4 and g(x) =2x^2+x-3, find the average rate of change of f(g(2x+8)) on the interval [-1,4]

Mathematics

1 answer:

avanturin [10]3 years ago

6 0

The average value of $f(g(2x+8))$ on the interval [-1, 4] is given by

$\dfrac{f(g(2\cdot4+8))-f(g(2\cdot(-1)+8))}{4-(-1)}=\dfrac{f(g(16))-f(g(6))}5$

We have

$g(16)=2\cdot16^2+16-3=525$

$f(525)=525^2-6\cdot525+4=272479$

$g(6)=2\cdot6^2+6-3=75$

$f(75)=5179$

$\implies\dfrac{f(g(16))-f(g(6))}5=53460$

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1 year ago

Which value of x is a solution of the inequality?

Goryan [66]

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4 years ago

Find the absolute maximum and absolute minimum values of f on the given interval. f(t) = 2 cos(t) + sin(2t),[0, π/2].

castortr0y [4]

Answer:

The absolute maximum is $\frac{3\sqrt 3}2$ and the absolute minimum value is $0.$

Step-by-step explanation:

Differentiate of $f$ both sides w.r.t. $t$ ,

$f(t)=2 \cos t+\sin 2t$

$\Rightarrow f'(t)=-2\sin t+2\cos 2t$

Now take $f'(t)=0$

$\Rightarrow -2\sin t+2\cos 2t=0$

$\Rightarrow 2\cos 2t=2\sin t$

$\Rightarrow \cos 2t=\sin t$

$\Rightarrow 1-2\sin ^2t =\sin t \quad \quad [\because \cos 2t = 1-2\sin ^2t]$

$\Rightarrow 2\sin ^2t+\sin t-1=0$

$\Rightarrow 2\sin ^2t+2\sin t-\sin t-1=0$

$\Rightarrow 2\sin t(\sin t+1)-1(\sin t+1)=0$

$\Rightarrow (\sin t+1)(2\sin t-1)=0$

$\Rightarrow \sin t+1=0 \;\text{and}\; 2\sin t-1=0$

$\Rightarrow \sin t =-1 \;\text{and}\; \sin t =\frac 12$

In the interval $0\leq t\leq \frac {\pi}2$ , the answer to this problem is $\frac {\pi}6$

Now find the second derivative of $f(t)$ w.r.t. $t$ ,

$f''(t)=-2\cos t-4\sin 2t$

$\Rightarrow \left[f''(t)\right]_{t=\frac {\pi}6}=-2\times \frac {\sqrt 3}2-4\times \frac{\sqrt 3}2=-3\sqrt 3$

Thus, $f(t)$ is maximum at $t=\frac {\pi}6$ and minimum at $t=0$

$\left[f(t)\right]_{t=\frac {\pi}6}=2\times \frac {\sqrt 3}2+\frac{\sqrt 3}2=\frac{3\sqrt 3}2\;\text{and}\;\left[f(t)\right]_{t=\frac{\pi}2}= 2\times 0+0=0$

Hence, the absolute maximum is $\frac{3\sqrt 3}2$ and the absolute minimum value is $0$ .

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3 years ago

V=Xf; speed = wavelength x frequency

TEA [102]

I don’t understand your questions because it’s all connected together. Could you maybe separate the actual question in the comments for me to help?

4 0

3 years ago

The average test score of the class was an 80 and the standard deviation was 2. About 99% of the class would score between what

Alisiya [41]

Answer:

Answer: 78 and 82.

Step-by-step explanation:

We have been given that the average test score of the class was an 80 and the standard deviation was 2. We are asked to find two values between which 68% of class will score.

We know that in a normal distribution approximately 68% of the data falls within one standard deviation of the mean.

So 68% scores will lie within one standard deviation below and above mean that is

Upon substituting our given values, we will get

Therefore, about 68% of the class would score between 78 and 81 and option B is the correct choice.

Step-by-step explanation: Hope this helps :D

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3 years ago

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