Answer to Question #158445 in Calculus for geek

Question

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.

Answer 1

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

i) -13

ii) not exist

iii) $\sqrt{3}-18$

iv) -15

v) -21

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

Explanation :

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

2.

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

Transform $x^2-2 \ to \ x$

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

(b)

$If \ \ g(x) = \sqrt{x^2-9} \\ Difference \ quotient \\ \frac{g(x+h)-g(x)}{h}=\frac{\sqrt{(x+h)^{2}-9}-\sqrt{x^{2}-9}}{h} \\ If \ \ g(x) = \sqrt{x^2}-9 = |x| - 9 \\ \frac{g(x+h)-g(x)}{h}=\frac{|x+h|-9-|x|+9}{h}=\frac{|x+h|-|x|}{h}$

3.

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

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