For f(x) = 3x+ 1 and g(x) = x2 - 6, find (f+g)(x).

Mathematics

2 answers:

mylen [45]3 years ago

7 0

Answer:

Step-by-step explanation:

$(f+g)(x)=f(x)+g(x)\\\\\text{Substitute}\ f(x)=3x+1,\ g(x)=x^2-6:\\\\(f+g)(x)=(3x+1)+(x^2-6)\qquad\text{combine like terms}\\\\(f+g)(x)=x^2+3x+(1-6)\\\\(f+g)(x)=x^2+3x-5$

Elina [12.6K]3 years ago

5 0

Your answer would be option B. x+3x-5

You might be interested in

Use a graph in a (-2π, 2π, π/2) by (-3, 3, 1) viewing rectangle to complete the identity.

yaroslaw [1]

First, notice that:

$2\tan (\frac{x}{2})=2\cdot(\pm\sqrt[]{\frac{1-cosx}{1+\cos x})}$

And in the denominator we have:

$1+\tan ^2(\frac{x}{2})=1+\frac{1-\cos x}{1+\cos x}=\frac{1+cosx+1-\cos x}{1+cosx}=\frac{2}{1+\cos x}$

then, we have on the original expression:

$\begin{gathered} \frac{2\tan(\frac{x}{2})}{1+\tan^2(\frac{x}{2})}=\frac{2\cdot\pm\sqrt[]{\frac{1-\cos x}{1+cosx}}}{\frac{2}{1+\cos x}}=\frac{2\cdot(\pm\sqrt[]{1-cosx})\cdot(1+\cos x)}{2\cdot(\sqrt[]{1+cosx})} \\ =(\sqrt[]{1-\cos x})\cdot(\sqrt[]{1+\cos x})=\sqrt[]{(1-\cos x)(1+\cos x)} \\ =\sqrt[]{1-\cos^2x}=\sqrt[]{\sin^2x}=\sin x \end{gathered}$

therefore, the identity equals to sinx

8 0

1 year ago

A=<br> -4<br> 3<br> 3<br> 6<br> 4<br> 5<br> -3<br> -3<br> 4.

Art [367]

Answer:

Can you please rewrite this correctly so I can solve?

4 0

2 years ago

Please help!!

Advocard [28]

from the given answers there is no answer