Math Snap

$f(x)=\left\{\begin{array}{ll} x^{2} & \text { for } x \leq 0 \\ a x+b & \text { for } x>0 \end{array}\right.$ 49. The function $f$ above has an inverse function for which of the following values of $a$ and $b$ ? (A) $a=-1, b=-2$ (B) $a=-1, b=2$ (C) $a=0, b=-1$ (D) $a=1, b=-2$ (E) $a=1, b=2$

STEP 1

Assumptions 1. The function $f$ is defined piecewise with $f(x) = x^2$ for $x \leq 0$ and $f(x) = ax + b$ for $x > 0$. 2. We are looking for values of $a$ and $b$ such that $f$ has an inverse function. 3. For a function to have an inverse, it must be one-to-one (injective), meaning it passes the horizontal line test.

STEP 2

Analyze the first piece of the function, $f(x) = x^2$ for $x \leq 0$. This part of the function is one-to-one on the interval $(-\infty, 0]$ because it is decreasing and passes the horizontal line test.

STEP 3

Analyze the second piece of the function, $f(x) = ax + b$ for $x > 0$. This part of the function will be one-to-one if $a \neq 0$ because it will be either strictly increasing or strictly decreasing, and thus pass the horizontal line test.

STEP 4

Determine the conditions for the function to be continuous at $x=0$. For $f$ to have an inverse, it must also be continuous. $f(0^-) = f(0^+)$

STEP 5

Calculate $f(0^-)$, which is the value of the function as $x$ approaches $0$ from the left. $f(0^-) = 0^2 = 0$

STEP 6

Calculate $f(0^+)$, which is the value of the function as $x$ approaches $0$ from the right. $f(0^+) = a \cdot 0 + b = b$

STEP 7

Set $f(0^-)$ equal to $f(0^+)$ to find the condition for continuity at $x=0$. $0 = b$

STEP 8

Now we know that for the function to be continuous at $x=0$, $b$ must be equal to $0$.

STEP 9

Review the options given for $a$ and $b$. We see that none of the options have $b=0$, which means we cannot use continuity to directly find the answer. Instead, we must ensure that the function is one-to-one across the entire domain.

STEP 10

Since the function $f(x) = x^2$ for $x \leq 0$ is already one-to-one and increasing as $x$ approaches $0$ from the left, the function $f(x) = ax + b$ for $x > 0$ must be increasing as $x$ approaches $0$ from the right for $f$ to be one-to-one across the entire domain.

STEP 11

The slope $a$ must be positive for $f(x) = ax + b$ to be increasing. Therefore, we can eliminate any options where $a$ is not positive.

STEP 12

Eliminate options (A), (B), and (C) because they have $a \leq 0$.

STEP 13

We are now left with options (D) and (E), which have $a=1$. We need to determine which value of $b$ will ensure that the function is one-to-one across the entire domain.

STEP 14

Since $a=1$ for both remaining options, the function $f(x) = x + b$ for $x > 0$ must connect smoothly with $f(x) = x^2$ for $x \leq 0$ at $x=0$ to maintain the one-to-one property.

STEP 15

We have already established that $f(0^-) = 0$. For the function to be one-to-one, the derivative of $f(x)$ must not change sign at $x=0$.

STEP 16

Calculate the derivative of $f(x)$ for $x \leq 0$. $f'(x) = \frac{d}{dx} x^2 = 2x$

STEP 17

Evaluate the derivative at $x=0$ from the left. $f'(0^-) = 2 \cdot 0 = 0$

STEP 18

Calculate the derivative of $f(x)$ for $x > 0$. $f'(x) = \frac{d}{dx} (ax + b) = a$

STEP 19

Since $a=1$ for both remaining options, the derivative from the right at $x=0$ is $f'(0^+) = 1$.

STEP 20

The derivative from the left is $0$ and from the right is $1$, which means the function is increasing on both sides of $x=0$. This satisfies the condition for the function to be one-to-one.

STEP 21

Now we must ensure that the function is continuous at $x=0$ to be invertible. We know that $f(0^-) = 0$ and $f(0^+) = a \cdot 0 + b = b$.

STEP 22

Since $f(0^-) = f(0^+)$ for the function to be continuous, we must have $b = 0$. However, since none of the options have $b=0$, we must choose the option where the jump discontinuity at $x=0$ is minimized, as the function can still be one-to-one with a jump discontinuity.

STEP 23

Review the remaining options (D) and (E) to determine which value of $b$ results in the smallest jump discontinuity at $x=0$.

STEP 24

Option (D) has $b=-2$ and option (E) has $b=2$. Since we want the function to be increasing and connect as smoothly as possible at $x=0$, we prefer $b$ to be as close to $0$ as possible from above.

SOLUTION

Choose option (E) with $a=1$ and $b=2$ because it will result in the function being continuous from the right at $x=0$ and the function will be increasing, thus maintaining the one-to-one property. The function $f$ has an inverse function for the values of $a=1$ and $b=2$. Solution: (E) $a=1, b=2$
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 103, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}