Question

1 what value of x cannot be evaluated in the function r(x)
2 evaluate the following: g(5)=__ p(5)=__ r(-9)=__ f(-9)
3 find f(10)
4 what is the value of g(-3)
5 find h(-8)
answer choices: -16,15,129,41,-23,25,3,-1/3

Answer 1

Answer:

2x+4 for x=5

Just replace the variable "x" with "5":

f(5) = 2×5 + 4 = 14

Answer: f(5) = 14

More Examples

Here is a function:

f(x) = 1 − x + x2

Important! The "x" is just a place-holder! And "f" is just a name.

These are all the same function:

f(x) = 1 − x + x2

f(q) = 1 − q + q2

w(A) = 1 − A + A2

pumpkin(θ) = 1 − θ + θ2

Evaluate For a Given Value:

Let us evaluate that function for x=3:

f(3) = 1 − 3 + 32 = 1 − 3 + 9 = 7

Evaluate For a Given Expression:

Evaluating can also mean replacing with an expression (such as 3m+1 or v2).

Let us evaluate the function for x=1/r:

f(1/r) = 1 − (1/r) + (1/r)2

Or evaluate the function for x = a−4:

f(a−4) = 1 − (a−4) + (a−4)2

= 1 − a + 4 + a2 − 8a + 16

= 21 − 9a + a2

Another Example

You can use your ability to evaluate functions to find other answers:

Example: h(x) = 3x2 + ax − 1

You are told that h(3) = 8, can you work out what "a" is?

First, evaluate h(3): h(3) = 3×(3)2 + a×3 − 1

Simplify: h(3) = 27 + 3a − 1

h(3) = 26 + 3a

Now ... we know that h(3) = 8, so: 8 = 26 + 3a

Swap sides: 26 + 3a = 8

Subtract 26 from both sides: 3a = −18

Divide by 3: a = −6

Check: h(3) = 3(3)2 − 6×3 − 1 = 27 − 18 − 1 = 8 yes

Careful!

I recommend putting the substituted values inside parentheses () , so you don't make mistakes.

Example: evaluate the function h(x) = x2 + 2 for x = −3

Replace the variable "x" with "−3":

h(−3) = (−3)2 + 2 = 9 + 2 = 11

Without the () you could make a mistake:

h(−3) = −32 + 2 = −9 + 2 = −7 (WRONG!)

Also be careful of this:

f(x+a) is not the same as f(x) + f(a)

Example: g(x) = x2

g(w+1) = (w+1)2 = w2 + 2w + 1

vs

g(w) + g(1) = w2 + 12 = w2 + 1

Answer 2

Answer:

Solution-:

Step-by-step explanation:

2x+4 for x=5

Just replace the variable "x" with "5":

f(5) = 2×5 + 4 = 14

Answer: f(5) = 14

More Examples

Here is a function:

f(x) = 1 − x + x2

Important! The "x" is just a place-holder! And "f" is just a name.

These are all the same function:

f(x) = 1 − x + x2

f(q) = 1 − q + q2

w(A) = 1 − A + A2

pumpkin(θ) = 1 − θ + θ2

Evaluate For a Given Value:

Let us evaluate that function for x=3:

f(3) = 1 − 3 + 32 = 1 − 3 + 9 = 7

Evaluate For a Given Expression:

Evaluating can also mean replacing with an expression (such as 3m+1 or v2).

Let us evaluate the function for x=1/r:

f(1/r) = 1 − (1/r) + (1/r)2

Or evaluate the function for x = a−4:

f(a−4) = 1 − (a−4) + (a−4)2

= 1 − a + 4 + a2 − 8a + 16

= 21 − 9a + a2

Another Example

You can use your ability to evaluate functions to find other answers:

Example: h(x) = 3x2 + ax − 1

You are told that h(3) = 8, can you work out what "a" is?

First, evaluate h(3): h(3) = 3×(3)2 + a×3 − 1

Simplify: h(3) = 27 + 3a − 1

h(3) = 26 + 3a

Now ... we know that h(3) = 8, so: 8 = 26 + 3a

Swap sides: 26 + 3a = 8

Subtract 26 from both sides: 3a = −18

Divide by 3: a = −6

Check: h(3) = 3(3)2 − 6×3 − 1 = 27 − 18 − 1 = 8 yes

Careful!

I recommend putting the substituted values inside parentheses () , so you don't make mistakes.

Example: evaluate the function h(x) = x2 + 2 for x = −3

Replace the variable "x" with "−3":

h(−3) = (−3)2 + 2 = 9 + 2 = 11

Without the () you could make a mistake:

h(−3) = −32 + 2 = −9 + 2 = −7 (WRONG!)

Also be careful of this:

f(x+a) is not the same as f(x) + f(a)

Example: g(x) = x2

g(w+1) = (w+1)2 = w2 + 2w + 1

vs

g(w) + g(1) = w2 + 12 = w2 + 1