Answer to Question #158445 in Calculus for geek

Question #158445

1. Let f(x) = √ x + 1 − 1 4−x2  . Compute the following: (10 points)

i. f(0)

ii. f(-3)

iii. f(2)

iv. f(-1)

v. f(3)

vi. Domain(f(x))


2. (a) Let f( x x−2 ) = 3x + 4 find f(x).

(b) Compute difference quotient of the function g(x) = √ x2 − 9, and simplify your answer.


3. Let f(x) = 1 − x, g(x) = x2 + bx + c. find b and c such that fog(x) = −x2  + 5x + 4. + bx + c. find b and c such that fog(x) = −x2+ 5x + 4.


1
Expert's answer
2021-01-29T12:45:51-0500

1. f(x)=x+114x2f(x) = \sqrt{x+1} -14-x^2 \\ \\

i) -13

ii) not exist

iii) 318\sqrt{3}-18

iv) -15

v) -21

vi) [1;+)[-1;+\infty)

Explanation :

 i) f(0)=0+11402=114=13 ii) f(3)=3114(3)2=4149= - not defined) because 4 is not defined iii) f(2)=2+114(2)2=3144=318 iv) f(1)=1+114(1)2=0141=15 v) f(3)=3+11932=2149=21 vi) x+1 needs x+10 , so domain is x[1;+)\text { i) } f(0)=\sqrt{0+1}-14-0^{2}=1-14=-13 \\ \text { ii) } f(-3)=\sqrt{-3-1}-14-(-3)^{2}=\sqrt{-4}-14-9= \\ \text { - not defined) because } \sqrt{-4} \text { is not defined} \\ \text { iii) } f(2)=\sqrt{2+1}-14-(2)^{2}=\sqrt{3}-14-4=\sqrt{3}-18 \\ \text { iv) } f(-1)=\sqrt{-1+1}-14-(-1)^{2}=0-14-1=-15 \\ \text { v) } f(3)=\sqrt{3+1}-19-3^{2}=2-14-9=-21 \\ \text { vi) } \sqrt{x+1} \text { needs } x+1 \geqslant 0 \text { , so domain is } x \in[-1 ;+\infty)


2.

(a) f(x22)=3x+4f(x^2-2) = 3x+4 \\

Transform x22 to xx^2-2 \ to \ x

Let t=x22,so x=±t+2f(t)=3x+4f(x)=3(±x+2)+4=±3x+2+4Let \ t=x^2-2, \quad so \ x = \pm\sqrt{t+2} \\ f(t) = 3x+4 \rightarrow f(x) = 3(\pm\sqrt{x+2}) + 4 = \pm3\sqrt{x+2} +4

(b)

If  g(x)=x29Difference quotientg(x+h)g(x)h=(x+h)29x29hIf  g(x)=x29=x9g(x+h)g(x)h=x+h9x+9h=x+hxhIf \ \ g(x) = \sqrt{x^2-9} \\ Difference \ quotient \\ \frac{g(x+h)-g(x)}{h}=\frac{\sqrt{(x+h)^{2}-9}-\sqrt{x^{2}-9}}{h} \\ If \ \ g(x) = \sqrt{x^2}-9 = |x| - 9 \\ \frac{g(x+h)-g(x)}{h}=\frac{|x+h|-9-|x|+9}{h}=\frac{|x+h|-|x|}{h}

3.

f(x)=1x,g(x)=x2+bx+c(fg)(x)=x2+5x+4f(x)=1-x, g(x)=x^{2}+b x+c \\ (f \circ g)(x)=-x^{2}+5 x+4 \\

(fg)(x)=f(g(x))=f(x2+bx+c)=1(x2+bx+c)==1x2bxc=x2bx+(1c)=x2+5x+4b=5,1c=4b=5,c=3\begin{array}{l} (f \circ g)(x)=f(g(x))=f\left(x^{2}+b x+c\right)=1-\left(x^{2}+b x+c\right)= \\ =1-x^{2}-b x-c=-x^{2}-bx+(1-c)=-x^{2}+5 x+4 \\ \Rightarrow-b=5,1-c=4 \Rightarrow b=-5, c=-3 \end{array}


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