Resources

Resources

Bytelearn - cat image with glasses

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

A function $g(x)$ increases by a factor of $7$ over every unit interval in $x$ and $g(0) = 1$ . $\newline$ Which could be a function rule for $g(x)$ ? $\newline$ Choices: $\newline$ (A) $g(x) = 7^x$ $\newline$ (B) $g(x) = 1.07^x$ $\newline$ (C) $g(x) = 1 + \frac{1}{7^x}$ $\newline$ (D) $g(x)$ = $1$ + $7x$

View step-by-step help

View step-by-step help

Intermediate Value Theorem

Full solution

Q. A function

g(x)

increases by a factor of

7

over every unit interval in

x

and

g(0) = 1

.

\newline

Which could be a function rule for

g(x)

?

\newline

Choices:

\newline

(A)

g(x) = 7^x

\newline

(B)

g(x) = 1.07^x

\newline

(C)

g(x) = 1 + \frac{1}{7^x}

\newline

(D)

g(x)

=

1

+

7x

Check $g(0)$ : Check $g(0)$ for each function to see if it equals $1$ . $\newline$ For (A) $g(0) = 7^0 = 1$ . $\newline$ For (B) $g(0) = 1.07^0 = 1$ . $\newline$ For (C) $g(0) = 1 + \frac{1}{7^0} = 1 + \frac{1}{1} = 2$ . $\newline$ For (D) $g(0) = 1 + 7\cdot0 = 1$ . $\newline$ Options (A), (B), and (D) pass this check, but (C) does not.
Check increase by factor: Check the increase by a factor of $7$ over one unit interval for the remaining options. $\newline$ For (A) $g(1) = 7^1 = 7$ , which is $7$ times $g(0)$ . $\newline$ For (B) $g(1) = 1.07^1 = 1.07$ , which is not $7$ times $g(0)$ . $\newline$ For (D) $g(1) = 1 + 7 \times 1 = 8$ , which is not $7$ times $g(0)$ . $\newline$ Only option (A) shows the correct increase by a factor of $7$ .
Final function rule: Since option (A) satisfies both conditions, $g(x) = 7^x$ is the correct function rule.

More problems from Intermediate Value Theorem

Question

Which of the following functions are continuous for all real numbers?

\newline

g(x)=\ln (x)

\newline

f(x)=\frac{1}{x}

\newline

1

\newline

g

\newline

f

\newline

g

f

\newline

g

f

Posted 3 months ago

Question

\newline

f(x)=\left\{\begin{array}{ll} \ln (-x)+3 & \text { for } x<-3 \\ \ln (-x+3) & \text { for }-3 \leq x<3 \end{array}\right.

\newline

f

x=-3

\newline

1

\newline

\newline

\mathrm{No}

Posted 2 months ago

Question

\newline

f(x)=\left\{\begin{array}{ll}\ln (x) & \text { for } 0<x \leq 2 \\ x^{2} \ln (2) & \text { for } x>2\end{array}\right.

\newline

f

x=2

\newline

1

\newline

\newline

\mathrm{No}

Posted 3 months ago

Question

\newline

g(x)=\left\{\begin{array}{ll} \frac{1}{\cos (x)} & \text { for }-\frac{\pi}{2}<x<0 \\ \cos (x+\pi) & \text { for } x \geq 0 \end{array}\right.

\newline

g

x=0

\newline

1

\newline

\newline

Posted 3 months ago

Question

\newline

h(x)=\left\{\begin{array}{ll} 2^{x-1} & \text { for } x<1 \\ 2^{1-x} & \text { for } x \geq 1 \end{array}\right.

\newline

h

x=1

\newline

1

\newline

\newline

Posted 3 months ago

Question

\newline

g(x)=\left\{\begin{array}{ll} (x-2)^{2} & \text { for } x \leq 2 \\ 2-x^{2} & \text { for } x>2 \end{array}\right.

\newline

g

x=2

\newline

1

\newline

\newline

\mathrm{No}

Posted 3 months ago

Question

\newline

g(x)=\left\{\begin{array}{ll} e^{x} & \text { for } x \leq-1 \\ -e^{-x} & \text { for } x>-1 \end{array}\right.

\newline

g

x=-1

\newline

1

\newline

\newline

Posted 3 months ago

Question

\newline

h(x)=\left\{\begin{array}{ll} e^{2 x} & \text { for } x<0 \\ e^{5 x} & \text { for } x \geq 0 \end{array}\right.

\newline

h

x=0

\newline

1

\newline

\newline

Posted 3 months ago

Question

\newline

h(x)=\left\{\begin{array}{cl}\frac{6}{x+5} & \text { for }-5<x<-3 \\ x^{2}-6 & \text { for } x \geq-3\end{array}\right.

\newline

h

x=-3

\newline

1

\newline

\newline

Posted 3 months ago

Question

\newline

g(x)=\left\{\begin{array}{ll}\sin (x) & \text { for } x<0 \\ x^{2} & \text { for } x \geq 0\end{array}\right.

\newline

g

x=0

\newline

1

\newline

\newline

Posted 3 months ago