What is the range of f(x)=1+sqrt(3+2x-x^2) ?

The range of f(x)=1+sqrt(3+2x-x^2) is 1<= f(x)<= 3

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range of f(x)=1+sqrt(3+2x-x^2)

range

f (x) = 1 + 3 + 2 x - x^{2}

1 \leq f (x) \leq 3

Interval Notation

[1, 3]

Solution steps

Find the minimum and maximum value in each defined interval and unite the results

Domain of

1 + 3 + 2 x - x^{2} : - 1 \leq x \leq 3

Extreme Points of

1 + 3 + 2 x - x^{2} :

Minimum

(- 1, 1),

Maximum

(1, 3),

Minimum

(3, 1)

Find the range for the interval

- 1 \leq x \leq 3 : 1 \leq f (x) \leq 3

Combine the ranges of all domain intervals to obtain the function range

1 \leq f (x) \leq 3

f (x) = \frac{x + 3}{x ^{2} - 49}

f (x) = - 5 x^{5} - 2 x^{3} + 4 x

f (x) = \frac{3 x - 2}{x + 5}

x + 3 y = 10

m = \frac{4}{1}, (- 9, - 10)

What is the range of f(x)=1+sqrt(3+2x-x^2) ?
The range of f(x)=1+sqrt(3+2x-x^2) is 1<= f(x)<= 3