Math / Algebra

QuestionIf $f(x)=x^{2}+4$ and $p(x)=\sqrt{x-1}$ , find $(f \circ p)(-6)$ . $(f \circ p)(-6)=$ $\square$ (Simplify your answer. Type an exact answer, using radicals

Studdy Solution

STEP 1

1. We are given two functions: $f(x) = x^2 + 4$ and $p(x) = \sqrt{x-1}$ .
2. We need to find the value of the composition of these functions, $(f \circ p)(-6)$ .

STEP 2

1. Understand the composition of functions $(f \circ p)(x)$ .
2. Evaluate $p(-6)$ .
3. Determine if $p(-6)$ is defined.
4. Substitute the result of $p(-6)$ into $f(x)$ .
5. Simplify the expression to find $(f \circ p)(-6)$ .

STEP 3

Understand the composition of functions:
$(f \circ p)(x) = f(p(x))$
This means we first apply $p(x)$ , then apply $f(x)$ to the result.

STEP 4

Evaluate $p(-6)$ :
$p(-6) = \sqrt{-6 - 1}$ $p(-6) = \sqrt{-7}$

STEP 5

Determine if $p(-6)$ is defined:
Since $\sqrt{-7}$ involves taking the square root of a negative number, $p(-6)$ is not defined in the real number system.

STEP 6

Since $p(-6)$ is not defined, $(f \circ p)(-6)$ is also not defined.
The value of $(f \circ p)(-6)$ is:
$\boxed{\text{undefined}}$

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QuestionIf $f(x)=x^{2}+4$ and $p(x)=\sqrt{x-1}$ , find $(f \circ p)(-6)$ . $(f \circ p)(-6)=$ $\square$ (Simplify your answer. Type an exact answer, using radicals

Studdy Solution

STEP 1

1. We are given two functions: $f(x) = x^2 + 4$ and $p(x) = \sqrt{x-1}$ .
2. We need to find the value of the composition of these functions, $(f \circ p)(-6)$ .

STEP 2

1. Understand the composition of functions $(f \circ p)(x)$ .
2. Evaluate $p(-6)$ .
3. Determine if $p(-6)$ is defined.
4. Substitute the result of $p(-6)$ into $f(x)$ .
5. Simplify the expression to find $(f \circ p)(-6)$ .

STEP 3

Understand the composition of functions:
$(f \circ p)(x) = f(p(x))$
This means we first apply $p(x)$ , then apply $f(x)$ to the result.

STEP 4

Evaluate $p(-6)$ :
$p(-6) = \sqrt{-6 - 1}$ $p(-6) = \sqrt{-7}$

STEP 5

Determine if $p(-6)$ is defined:
Since $\sqrt{-7}$ involves taking the square root of a negative number, $p(-6)$ is not defined in the real number system.

STEP 6

Since $p(-6)$ is not defined, $(f \circ p)(-6)$ is also not defined.
The value of $(f \circ p)(-6)$ is:
$\boxed{\text{undefined}}$

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QuestionIf f(x)=x2+4f(x)=x^{2}+4f(x)=x2+4 and p(x)=x−1p(x)=\sqrt{x-1}p(x)=x−1​, find (f∘p)(−6)(f \circ p)(-6)(f∘p)(−6). (f∘p)(−6)=(f \circ p)(-6)=(f∘p)(−6)= □\square□ (Simplify your answer. Type an exact answer, using radicals

Studdy Solution

STEP 1

1. We are given two functions: f(x)=x2+4 f(x) = x^2 + 4 f(x)=x2+4 and p(x)=x−1 p(x) = \sqrt{x-1} p(x)=x−1​.2. We need to find the value of the composition of these functions, (f∘p)(−6) (f \circ p)(-6) (f∘p)(−6).

STEP 2

STEP 3

Understand the composition of functions:(f∘p)(x)=f(p(x)) (f \circ p)(x) = f(p(x)) (f∘p)(x)=f(p(x))This means we first apply p(x) p(x) p(x), then apply f(x) f(x) f(x) to the result.

STEP 4

Evaluate p(−6) p(-6) p(−6):p(−6)=−6−1 p(-6) = \sqrt{-6 - 1} p(−6)=−6−1​ p(−6)=−7 p(-6) = \sqrt{-7} p(−6)=−7​

STEP 5

Determine if p(−6) p(-6) p(−6) is defined:Since −7\sqrt{-7}−7​ involves taking the square root of a negative number, p(−6) p(-6) p(−6) is not defined in the real number system.

STEP 6

Since p(−6) p(-6) p(−6) is not defined, (f∘p)(−6) (f \circ p)(-6) (f∘p)(−6) is also not defined. The value of (f∘p)(−6) (f \circ p)(-6) (f∘p)(−6) is:undefined \boxed{\text{undefined}} undefined​

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QuestionIf $f(x)=x^{2}+4$ and $p(x)=\sqrt{x-1}$ , find $(f \circ p)(-6)$ . $(f \circ p)(-6)=$ $\square$ (Simplify your answer. Type an exact answer, using radicals

1. We are given two functions: $f(x) = x^2 + 4$ and $p(x) = \sqrt{x-1}$ .
2. We need to find the value of the composition of these functions, $(f \circ p)(-6)$ .

Understand the composition of functions:
$(f \circ p)(x) = f(p(x))$
This means we first apply $p(x)$ , then apply $f(x)$ to the result.

Evaluate $p(-6)$ :
$p(-6) = \sqrt{-6 - 1}$ $p(-6) = \sqrt{-7}$

Determine if $p(-6)$ is defined:
Since $\sqrt{-7}$ involves taking the square root of a negative number, $p(-6)$ is not defined in the real number system.

Since $p(-6)$ is not defined, $(f \circ p)(-6)$ is also not defined.
The value of $(f \circ p)(-6)$ is:
$\boxed{\text{undefined}}$