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3 years ago

Which are correct representations of the inequality -3 (2x - 5) < 5 (2 - x)? Check all that apply

Mathematics

1 answer:

viva [34]3 years ago

4 0

Im not sure if this is correct but I got that -3(2x-5)<5(2-x) so I got the same thing as the question

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3 years ago

Find (f −1)'(a).<br><br> f(x) = 4 + x2 + tan(πx/2), −1 < x < 1, a = 4

Vinvika [58]

Answer:

The solution is $(f^{-1})'(a) = \frac{2}{\pi }$

Step-by-step explanation:

From the question we are told that

The function is $f(x) = 4 + x^2 + tan [\frac{ \pi x}{2} ]$ , -1 < x < 1 a = 4

Here we are told find $(f^{-1}) (a)$

Let equate

$f(x) = a$

$4 + x^2 + tan[\frac{\pi x }{2} ] = 4$

$x^2 + tan[\frac{\pi x }{2} ] = 0$

For the equation above to be valid x must be equal to 0

Now when x = 0

$f(0) = 4+0^2 + tan [\frac{ \pi * 0}{2} ]$

=> $f(0) = 4$

=> $0 = f^{-1} (4)$

Differentiating f(x)

$f(x)' = 0 + 2x + sec^2 (\frac{\pi x}{2} )\cdot \frac{\pi}{2}$

Now

since $0 = f^{-1} (4)$

We have

$f(0)' = 0 + \frac{\pi }{2} sec^2 (0)$

$f(0)' = \frac{\pi }{2}$

Now

$(f^{-1})'(a) = \frac{1}{(\frac{\pi}{2} )}$

$(f^{-1})'(a) = \frac{2}{\pi }$

3 0

3 years ago