Let f(x)=2x+3 and g(x)=x×x-1find a. f(g(x)) b.g(f(x))

Mathematics

1 answer:

Talja [164]2 years ago

5 0

Answer:

See explanations below

Step-by-step explanation:

Given the functions f(x)=2x+3 and g(x)=x^2-1

a. Find f(g(x))

f(g(x)) = f(x^2-1)

f(g(x)) = 2(x^2-1)+3

f(g(x))= 2x^2-2+3

f(g(x)) = 2x^2+1

Hence the composite function f(g(x)) is 2x^2+1

b) g(f(x)) = g(2x+3)

g(f(x) = (2x+3)^2-1

g(f(x)) = 4x^2+12x+9-1

g(f(x)) = 4x^2+12x+8

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Suppose cos(x)= -1/3, where π/2 ≤ x ≤ π. What is the value of tan(2x). EDGE

AVprozaik [17]

Answer:

Step-by-step explanation:

We are given that:

$\displaystyle \cos x = -\frac{1}{3}\text{ where } \pi /2 \leq x \leq \pi$

And we want to find the value of tan(2x).

Note that since x is between π/2 and π, it is in QII.

In QII, cosine and tangent are negative and only sine is positive.

We can rewrite our expression as:

$\displaystyle \tan(2x)=\frac{\sin(2x)}{\cos(2x)}$

Using double angle identities:

$\displaystyle \tan(2x)=\frac{2\sin x\cos x}{\cos^2 x-\sin^2 x}$

Since cosine relates the ratio of the adjacent side to the hypotenuse and we are given that cos(x) = -1/3, this means that our adjacent side is one and our hypotenuse is three (we can ignore the negative). Using this information, find the opposite side:

$\displaystyle o=\sqrt{3^2-1^2}=\sqrt{8}=2\sqrt{2}$

So, our adjacent side is 1, our opposite side is 2√2, and our hypotenuse is 3.

From the above information, substitute in appropriate values. And since x is in QII, cosine and tangent will be negative while sine will be positive. Hence: