Let h(x)={[sqrt(-x+1)," for "x = 1]:} Is h continuous at x=1 ? Choose 1 answer: (A) Yes (B) No

Check Conditions: To determine if h(x) is continuous at x=1, we need to check if the following three conditions are met: 1. h(1) is defined. 2. The limit of h(x) as x approaches 1 from the left (lim x->1-) is equal to h(1). 3. The limit of h(x) as x approaches 1 from the right (lim x->1+) is equal to h(1).Find h(1): First, we find h(1) using the definition of h(x) for x >= 1, which is h(x) = sqrt(2x). Substitute x = 1 into the function to get h(1) = sqrt(2*1) = sqrt(2).Limit from Left: Next, we find the limit of h(x) as x approaches 1 from the left, which is the limit of sqrt(-x+1) as x approaches 1. lim x->1- sqrt(-x+1) = sqrt(-(1)+1) = sqrt(0) = 0.Limit from Right: Now, we find the limit of h(x) as x approaches 1 from the right, which is the limit of sqrt(2x) as x approaches 1. lim x->1+ sqrt(2x) = sqrt(2*1) = sqrt(2).Compare Limits: We compare the limit from the left and the limit from the right to the value of h(1). The limit from the left is 0, and the limit from the right and the value of h(1) are both sqrt(2). Since the limits from both sides do not match, h(x) is not continuous at x=1.

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Let

h(x)={[sqrt(-x+1)," for "x < 1],[sqrt(2x)," for "x >= 1]:}
Is
h continuous at
x=1 ?
Choose 1 answer:
(A) Yes
(B) No

Let $\newline$ $h(x)=\left\{\begin{array}{ll} \sqrt{-x+1} & \text { for } x<1 \\ \sqrt{2 x} & \text { for } x \geq 1 \end{array}\right.$ $\newline$ Is $h$ continuous at $x=1$ ? $\newline$ Choose $1$ answer: $\newline$ (A) Yes $\newline$ (B) No

View step-by-step help

Home

Math Problems

Calculus

Intermediate Value Theorem

Full solution

Q. Let

\newline

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

\newline

h

continuous at

x=1

\newline

Choose

1

answer:

\newline

(A) Yes

\newline

(B) No

Check Conditions: To determine if $h(x)$ is continuous at $x=1$ , we need to check if the following three conditions are met: $\newline$ $1$ . $h(1)$ is defined. $\newline$ $2$ . The limit of $h(x)$ as $x$ approaches $1$ from the left ( $\lim_{x\to1-}$ ) is equal to $h(1)$ . $\newline$ $3$ . The limit of $h(x)$ as $x$ approaches $1$ from the right ( $x=1$ $1$ ) is equal to $h(1)$ .
Find $h(1)$ : First, we find $h(1)$ using the definition of $h(x)$ for $x \geq 1$ , which is $h(x) = \sqrt{2x}$ . Substitute $x = 1$ into the function to get $h(1) = \sqrt{2 \times 1} = \sqrt{2}$ .
Limit from Left: Next, we find the limit of $h(x)$ as $x$ approaches $1$ from the left, which is the limit of $\sqrt{-x+1}$ as $x$ approaches $1$ . $\newline$ $\lim_{x\to 1^-} \sqrt{-x+1} = \sqrt{-(1)+1} = \sqrt{0} = 0.$
Limit from Right: Now, we find the limit of $h(x)$ as $x$ approaches $1$ from the right, which is the limit of $\sqrt{2x}$ as $x$ approaches $1$ . $\lim_{x\to 1^+} \sqrt{2x} = \sqrt{2\cdot 1} = \sqrt{2}$ .
Compare Limits: We compare the limit from the left and the limit from the right to the value of $h(1)$ . The limit from the left is $0$ , and the limit from the right and the value of $h(1)$ are both $\sqrt{2}$ . Since the limits from both sides do not match, $h(x)$ is not continuous at $x=1$ .