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3 years ago

8

What is the slope intercept form

1 answer:

pochemuha3 years ago

7 0

Answer:

$y = \frac{4}{3} x - 4$

Hope it helped:)

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$f(x) = e^x sin (x)$

To find increasing and decreasing intervals we take derivative

$f'(x) = e^xsin(x)+e^x(cosx)= e^x(sinx+cosx)$

Now we set the derivative =0 and solve for x

$e^x(sinx+cosx)=0$

sinx + cosx =0

divide whole equation by cos x

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tanx +1 =0

tanx = 1

$x=\frac{3\pi }{4}$ and $x=\frac{7\pi}{4}$

Now we pick a number between 0 to $\frac{3\pi }{4}$

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Plug it into the derivative

$f'(x) =e^{\frac{\pi }{2}}(sin(\frac{\pi}{2})+cos(\frac{\pi }{2}))$

= 4.810 is positive

So the graph of f(x) is increasing on the interval [0, $x=\frac{3\pi }{4}$ )

Now we pick a number between $\frac{7\pi}{4}$ to 2pi

Lets pick $\frac{11\pi}{6}$

Plug it into the derivative

$f'(x) =e^{\frac{11\pi}{6}}(sin(\frac{11\pi}{6})+cos(\frac{11\pi }{6}))$

= 116 is positive

So the graph of f(x) is increasing on the interval $(\frac{7\pi }{4}, 2\pi)$

Increasing interval is $(0,\frac{3\pi }{4}) U (\frac{7\pi }{4}, 2\pi)$

Decreasing interval is $(\frac{3\pi}{4}, \frac{7\pi}{4})$

(b)

The graph of f(x) increases and reaches a local maximum at $x=\frac{3\pi}{4}$

The graph of f(x) decreases and reaches a local minimum at $x=\frac{7\pi}{4}$

(c)

f(0) = 0

$f(2\pi)=0$

$f(\frac{3\pi }{4})=7.46$

$f(\frac{7\pi}{4})=-172.64$

Here global maximum at $x=\frac{3\pi}{4}$

Here global minimum at $x=\frac{7\pi}{4}$

3 0

3 years ago

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