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已知函数 $f(x)=\left\{\begin{array}{l}-\frac{1}{x}+2, x

Mar 8, 2024

已知函数

f(x)=\left\{\begin{array}{l}-\frac{1}{x}+2, x&lt;c \\ x^{2}-2 x+3, c \leq x \leq 3\end{array}\right.

，若

f(x)

的值域为

[2,6]

, 则实数

c

的取值范围是 ( ) A.

\left[-1,-\frac{1}{4}\right]

\left[-\frac{1}{4}, 0\right)

[-1,0)

\left[-1,-\frac{1}{2}\right]

Generated Graph

Solution by Steps

step 1

To find the range of

c

for the function

f(x)

, we need to consider the value ranges of both pieces of the piecewise function separately

step 2

For x < c ,

f(x) = -\frac{1}{x} + 2

. The asksia-ll calculator indicates that there are no solutions for

x

when solving

-\frac{1}{x} + 2 = 2

, which means the function does not touch the value 2

step 3

When solving

-\frac{1}{x} + 2 = 6

, the asksia-ll calculator gives

x = -\frac{1}{4}

. This is the x-value where the function reaches the upper bound of the range

step 4

For

c \leq x \leq 3

f(x) = x^2 - 2x + 3

. The asksia-ll calculator finds that when solving

x^2 - 2x + 3 = 2

x = 1

, which is within the range of

c

to 3

step 5

Since

f(x)

must be within the range of

[2, 6]

c

must be chosen such that

f(x)

does not fall below 2

step 6

The quadratic part of

f(x)

is always greater than or equal to 2 for

x

in the interval

[c, 3]

, so we only need to consider the first part

-\frac{1}{x} + 2

for x < c

step 7

Since

-\frac{1}{x} + 2

reaches 6 when

x = -\frac{1}{4}

, and we want

f(x)

to be at least 2,

c

must be greater than or equal to

-\frac{1}{4}

step 8

However,

c

cannot be 0 because

-\frac{1}{x} + 2

would be undefined at

x = 0

. Therefore,

c

must be less than 0

step 9

Combining the conditions from steps 7 and 8, the range of

c

\left[-\frac{1}{4}, 0\right)

Answer

\left[-\frac{1}{4}, 0\right)

Key Concept

Piecewise Function Range Analysis

Explanation

To determine the range of

c

for the given piecewise function

f(x)

, we analyzed the function's behavior on both intervals separately and found the values of

c

that keep

f(x)

within the desired range of

[2, 6]