Math / Algebra

Question(26) If $f(x)=\left\{\begin{array}{ll}\mathrm{a} x^{2}-3, & x \neq 2 \\ 2 \mathrm{a} & , x=2\end{array}\right.$ is continuous at $x=2$ , then $\mathrm{a}=$ (a) $\frac{1}{2}$ (b) $\frac{2}{3}$ (c) $\frac{3}{2}$ (d) 6

Studdy Solution

STEP 1

What is this asking? We need to find the value of $a$ that makes the piecewise function $f(x)$ continuous at $x = 2$ . Watch out! Remember, for a function to be continuous at a point, the left-hand limit, the right-hand limit, and the function's value at that point must all be equal!

STEP 2

1. Define Continuity
2. Apply the Definition
3. Solve for a

STEP 3

Alright, so what does it mean for a function to be continuous at a point?
It means the function doesn't have any jumps or holes at that point!
More precisely, it means the limit of the function as $x$ approaches that point must be equal to the function's value at that point.

STEP 4

In our case, we're told $f(x)$ is continuous at $x = 2$ .
This means: $\lim_{x \to 2} f(x) = f(2)$

STEP 5

We know $f(2) = 2a$ .
Now we need to find the limit as $x$ approaches 2.
Since we're looking at the limit as $x$ approaches 2, and not at 2, we use the first part of the piecewise function: $ax^2 - 3$ .

STEP 6

So, we have: $\lim_{x \to 2} (ax^2 - 3) = 2a$

STEP 7

Let's substitute $x = 2$ into the expression $ax^2 - 3$ : $a \cdot 2^2 - 3 = 2a$ $4a - 3 = 2a$

STEP 8

Time to solve for $a$ !
Let's subtract $2a$ from both sides of the equation: $4a - 3 - 2a = 2a - 2a$ $2a - 3 = 0$

STEP 9

Now, let's add 3 to both sides: $2a - 3 + 3 = 0 + 3$ $2a = 3$

STEP 10

Finally, let's divide both sides by 2: $\frac{2a}{2} = \frac{3}{2}$ $a = \frac{3}{2}$

STEP 11

So, the value of $a$ that makes $f(x)$ continuous at $x = 2$ is $\frac{3}{2}$ , which corresponds to answer choice (c).

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Question(26) If $f(x)=\left\{\begin{array}{ll}\mathrm{a} x^{2}-3, & x \neq 2 \\ 2 \mathrm{a} & , x=2\end{array}\right.$ is continuous at $x=2$ , then $\mathrm{a}=$ (a) $\frac{1}{2}$ (b) $\frac{2}{3}$ (c) $\frac{3}{2}$ (d) 6

Studdy Solution

STEP 1

What is this asking? We need to find the value of $a$ that makes the piecewise function $f(x)$ continuous at $x = 2$ . Watch out! Remember, for a function to be continuous at a point, the left-hand limit, the right-hand limit, and the function's value at that point must all be equal!

STEP 2

1. Define Continuity
2. Apply the Definition
3. Solve for a

STEP 3

Alright, so what does it mean for a function to be continuous at a point?
It means the function doesn't have any jumps or holes at that point!
More precisely, it means the limit of the function as $x$ approaches that point must be equal to the function's value at that point.

STEP 4

In our case, we're told $f(x)$ is continuous at $x = 2$ .
This means: $\lim_{x \to 2} f(x) = f(2)$

STEP 5

We know $f(2) = 2a$ .
Now we need to find the limit as $x$ approaches 2.
Since we're looking at the limit as $x$ approaches 2, and not at 2, we use the first part of the piecewise function: $ax^2 - 3$ .

STEP 6

So, we have: $\lim_{x \to 2} (ax^2 - 3) = 2a$

STEP 7

Let's substitute $x = 2$ into the expression $ax^2 - 3$ : $a \cdot 2^2 - 3 = 2a$ $4a - 3 = 2a$

STEP 8

Time to solve for $a$ !
Let's subtract $2a$ from both sides of the equation: $4a - 3 - 2a = 2a - 2a$ $2a - 3 = 0$

STEP 9

Now, let's add 3 to both sides: $2a - 3 + 3 = 0 + 3$ $2a = 3$

STEP 10

Finally, let's divide both sides by 2: $\frac{2a}{2} = \frac{3}{2}$ $a = \frac{3}{2}$

STEP 11

So, the value of $a$ that makes $f(x)$ continuous at $x = 2$ is $\frac{3}{2}$ , which corresponds to answer choice (c).

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Question(26) If f(x)={ax2−3,x≠22a,x=2f(x)=\left\{\begin{array}{ll}\mathrm{a} x^{2}-3, & x \neq 2 \\ 2 \mathrm{a} & , x=2\end{array}\right.f(x)={ax2−3,2a​x=2,x=2​ is continuous at x=2x=2x=2, then a=\mathrm{a}=a= (a) 12\frac{1}{2}21​ (b) 23\frac{2}{3}32​ (c) 32\frac{3}{2}23​ (d) 6

Studdy Solution

STEP 1

What is this asking? We need to find the value of aaa that makes the piecewise function f(x)f(x)f(x) continuous at x=2x = 2x=2. Watch out! Remember, for a function to be continuous at a point, the left-hand limit, the right-hand limit, and the function's value at that point must all be equal!

STEP 2

1. Define Continuity2. Apply the Definition3. Solve for *a*

STEP 3

Alright, so what does it mean for a function to be continuous at a point?It means the function doesn't have any jumps or holes at that point!More precisely, it means the limit of the function as xxx approaches that point must be equal to the function's value at that point.

STEP 4

In our case, we're told f(x)f(x)f(x) is continuous at x=2x = 2x=2.This means: lim⁡x→2f(x)=f(2) \lim_{x \to 2} f(x) = f(2) x→2lim​f(x)=f(2)

STEP 5

We know f(2)=2af(2) = 2af(2)=2a.Now we need to find the limit as xxx approaches 2.Since we're looking at the limit as xxx approaches 2, and not *at* 2, we use the first part of the piecewise function: ax2−3ax^2 - 3ax2−3.

STEP 6

So, we have: lim⁡x→2(ax2−3)=2a \lim_{x \to 2} (ax^2 - 3) = 2a x→2lim​(ax2−3)=2a

STEP 7

Let's substitute x=2x = 2x=2 into the expression ax2−3ax^2 - 3ax2−3: a⋅22−3=2a a \cdot 2^2 - 3 = 2a a⋅22−3=2a 4a−3=2a 4a - 3 = 2a 4a−3=2a

STEP 8

**Time to solve for** aaa!Let's subtract 2a2a2a from both sides of the equation: 4a−3−2a=2a−2a 4a - 3 - 2a = 2a - 2a 4a−3−2a=2a−2a 2a−3=0 2a - 3 = 0 2a−3=0

STEP 9

Now, let's add 3 to both sides: 2a−3+3=0+3 2a - 3 + 3 = 0 + 3 2a−3+3=0+3 2a=3 2a = 3 2a=3

STEP 10

Finally, let's divide both sides by 2: 2a2=32 \frac{2a}{2} = \frac{3}{2} 22a​=23​ a=32 a = \frac{3}{2} a=23​

STEP 11

So, the value of aaa that makes f(x)f(x)f(x) continuous at x=2x = 2x=2 is 32\frac{3}{2}23​, which corresponds to answer choice (c).

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Question(26) If $f(x)=\left\{\begin{array}{ll}\mathrm{a} x^{2}-3, & x \neq 2 \\ 2 \mathrm{a} & , x=2\end{array}\right.$ is continuous at $x=2$ , then $\mathrm{a}=$ (a) $\frac{1}{2}$ (b) $\frac{2}{3}$ (c) $\frac{3}{2}$ (d) 6

What is this asking? We need to find the value of $a$ that makes the piecewise function $f(x)$ continuous at $x = 2$ . Watch out! Remember, for a function to be continuous at a point, the left-hand limit, the right-hand limit, and the function's value at that point must all be equal!

1. Define Continuity
2. Apply the Definition
3. Solve for a

Alright, so what does it mean for a function to be continuous at a point?
It means the function doesn't have any jumps or holes at that point!
More precisely, it means the limit of the function as $x$ approaches that point must be equal to the function's value at that point.

In our case, we're told $f(x)$ is continuous at $x = 2$ .
This means: $\lim_{x \to 2} f(x) = f(2)$

We know $f(2) = 2a$ .
Now we need to find the limit as $x$ approaches 2.
Since we're looking at the limit as $x$ approaches 2, and not at 2, we use the first part of the piecewise function: $ax^2 - 3$ .

So, we have: $\lim_{x \to 2} (ax^2 - 3) = 2a$

Let's substitute $x = 2$ into the expression $ax^2 - 3$ : $a \cdot 2^2 - 3 = 2a$ $4a - 3 = 2a$

Time to solve for $a$ !
Let's subtract $2a$ from both sides of the equation: $4a - 3 - 2a = 2a - 2a$ $2a - 3 = 0$

Now, let's add 3 to both sides: $2a - 3 + 3 = 0 + 3$ $2a = 3$

Finally, let's divide both sides by 2: $\frac{2a}{2} = \frac{3}{2}$ $a = \frac{3}{2}$

So, the value of $a$ that makes $f(x)$ continuous at $x = 2$ is $\frac{3}{2}$ , which corresponds to answer choice (c).