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3 years ago

Find (f • g) when f(x) = x^2 + 5x + 6 and g(x) = 1/x+3

Mathematics

2 answers:

Nikitich [7]3 years ago

8 0

Answer:

option D

Step-by-step explanation:

$f(x) = x^2 + 5x + 6$

$g(x)= \frac{1}{x+3}$

(fog)(x) = f(g(x))

Plug in g(x) in f(x)

We plug in 1/x+3 in the place of x in f(x)

$f(g(x))= f(\frac{1}{x+3})= (\frac{1}{x+3})^2 + 5(\frac{1}{x+3}) + 6$

To simplify it we take LCD

LCD is (x+3)(x+3)

$\frac{1}{(x+3)(x+3)}+5\frac{1*(x+3)}{(x+3)(x+3)}+\frac{6(x+3)(x+3)}{(x+3)(x+3)}$

$\frac{1}{x^2+6x+9}+\frac{(5x+15)}{x^2+6x+9}+\frac{6x^2+36x+54}{x^2+6x+9}$

All the denominators are same so we combine the numerators

$\frac{1+5x+15+6x^2+36x+54}{x^2+6x+9}$

$\frac{6x^2+41x+70}{x^2+6x+9}$

Option D is correct

spayn [35]3 years ago

4 0

Answer:

The correct option choice is D. 6x^2 + 41x + 70 / x^2 +6x + 9

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kirill [66]

Answer:

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Step-by-step explanation:

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liberstina [14]

Answer:

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Step-by-step explanation:

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Explanation:
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3 years ago

Derivative of tan(2x+3) using first principle

kodGreya [7K]

$f(x)=\tan(2x+3)$

The derivative is given by the limit

$f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h$

You have

$\displaystyle\lim_{h\to0}\frac{\tan(2(x+h)+3)-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan((2x+3)+2h)-\tan(2x+3)}h$

Use the angle sum identity for tangent. I don't remember it off the top of my head, but I do remember the ones for (co)sine.

$\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}=\dfrac{\sin a\cos b+\cos a\sin b}{\cos a\cos b-\sin a\sin b}=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$

By this identity, you have

$\tan((2x+3)+2h)=\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}$

So in the limit you get

$\displaystyle\lim_{h\to0}\frac{\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan(2x+3)+\tan2h-\tan(2x+3)(1-\tan(2x+3)\tan2h)}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h+\tan^2(2x+3)\tan2h}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h}h\times\lim_{h\to0}\frac{1+\tan^2(2x+3)}{1-\tan(2x+3)\tan2h}$
$\displaystyle\frac12\lim_{h\to0}\frac1{\cos2h}\times\lim_{h\to0}\frac{\sin2h}{2h}\times\lim_{h\to0}\frac{\sec^2(2x+3)}{1-\tan(2x+3)\tan2h}$

The first two limits are both 1, and the single term in the last limit approaches 0 as $h\to0$ , so you're left with

$f'(x)=\dfrac12\sec^2(2x+3)$

which agrees with the result you get from applying the chain rule.

7 0

3 years ago