Math  /  Algebra

Questionf(x)={c+cxx2 for x<37 for x=32c+3x2 for x>3f(x)=\left\{\begin{array}{ll} c+c x-x^{2} & \text { for } x<3 \\ 7 & \text { for } x=3 \\ 2 c+\frac{3}{x-2} & \text { for } x>3 \end{array}\right.
Let ff be the function defined above. For what value of cc, if any, is ff continuous at x=3x=3 ? (A) 2 (B) 4 (C) 6
D There is no such C .

Studdy Solution

STEP 1

What is this asking? We need to find a value for cc that makes this piecewise function connect smoothly at x=3x = 3. Watch out! Remember, for a function to be continuous at a point, the limit as xx approaches that point must be equal to the function's value *at* that point.

STEP 2

1. Find the left-hand limit.
2. Find the right-hand limit.
3. Set the limits equal to the function's value at x=3x = 3.

STEP 3

As xx approaches **3** from the *left*, we use the first piece of the function: c+cxx2c + cx - x^2.
Let's plug in x=3x = 3 and see what we get!

STEP 4

limx3f(x)=limx3(c+cxx2)=c+c332=c+3c9=4c9\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (c + cx - x^2) = c + c \cdot 3 - 3^2 = c + 3c - 9 = 4c - 9 So, the left-hand limit is 4c94c - 9.

STEP 5

As xx approaches **3** from the *right*, we use the third piece of the function: 2c+3x22c + \frac{3}{x-2}.
Let's substitute x=3x = 3 into this piece!

STEP 6

limx3+f(x)=limx3+(2c+3x2)=2c+332=2c+31=2c+3\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} \left(2c + \frac{3}{x-2}\right) = 2c + \frac{3}{3-2} = 2c + \frac{3}{1} = 2c + 3 The right-hand limit is 2c+32c + 3.

STEP 7

For f(x)f(x) to be continuous at x=3x = 3, the left-hand limit, the right-hand limit, and the function's value at x=3x = 3 must all be equal.
We know that f(3)=7f(3) = 7.

STEP 8

We found that the left-hand limit is 4c94c - 9 and the right-hand limit is 2c+32c + 3.
These must be equal to each other *and* equal to f(3)=7f(3) = 7.

STEP 9

Let's set the left-hand limit and the right-hand limit equal to each other: 4c9=2c+34c - 9 = 2c + 3 Subtract 2c2c from both sides: 2c9=32c - 9 = 3 Add 9 to both sides: 2c=122c = 12 Divide both sides by 2: c=6c = 6

STEP 10

Now, let's check if this value of cc makes both limits equal to f(3)=7f(3) = 7. Left-hand limit: 4(6)9=249=154(6) - 9 = 24 - 9 = 15 Right-hand limit: 2(6)+3=12+3=152(6) + 3 = 12 + 3 = 15 Since both limits equal **15**, but f(3)=7f(3) = 7, there's no value of cc that makes the function continuous at x=3x = 3.

STEP 11

There is no such value of cc.
The answer is (D).

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