Find the indicated function values. f(x)=(-x)^3-x^2-x+7 a. f(0) b. f(2) c. f(-2) d. f(1)+f(-1)

Name: Problem 17, Chapter 2: Introductory and Intermediate Algebra for College Students
Uploaded: 2020-06-28T21:17:01-08:00
Duration: 7 min 16 s
Channel: Sheila Akinleye
Description: Find the indicated function values. f(x)=(-x)^3-x^2-x+7 a. f(0) b. f(2) c. f(-2) d. f(1)+f(-1)

Find the indicated function values. f(x)=(-x)^3-x^2-x+7 a....

Key Concepts

Function Evaluation

This concept involves determining the output of a function for specific input values. Instead of dealing with the abstract definition of a function, evaluation replaces the independent variable with a given number to compute the corresponding result.

Substitution

Substitution is a fundamental technique in algebra where you replace the variables in an expression with specific numbers. This process is key to evaluating functions, as it lets you see how the function behaves for different inputs.

Exponentiation and Negative Bases

When a negative number is raised to a power, the parity of the exponent (whether it is even or odd) influences the result. In particular, for odd exponents the negative sign is preserved, while even exponents yield a positive result. This concept is crucial when evaluating functions with expressions like (-x) raised to an exponent.

Arithmetic Simplification

After substituting specific values into an algebraic expression, arithmetic simplification combines like terms and performs basic operations. This step is essential for obtaining the final, simplified numerical value of a function evaluation.

Transcript

00:02 Okay, so we have as our function here, f of x equal to negative x raised to the fifth power, minus x squared, minus x plus five.

00:24 Now, we want to find certain values of this function, and i'm going to first of all rewrite the function here in the top of the middle column because this will be our reference point.

00:38 And i'm emphasizing the fact that this middle column here is for our calculations.

00:44 And you do want to grab a calculator, have one handy.

00:48 So we're looking for the value of f of zero with this function, f of 3, f of negative 3, and then finally, f of 2 plus f of negative 2.

01:08 And we'll reserve all of the space down here for that part.

01:12 So i want to make the substitution first, which is to put zero in for x wherever i see it in my function.

01:20 That's going to be negative 0 to the fifth power, minus 0 to the fifth power, minus 0 plus 5.

01:29 So as you can probably guess, we just end up with 5.

01:32 When we do the same thing for 3, we're going to have, and notice how this is a positive 3.

01:38 So i'm just putting three in place of x negative three raised to the fifth minus three raised to the second minus three plus five now negative three raised to the fifth power is like saying negative three times negative three times negative three times negative three right we know that's going to be a negative number because an odd number for my x -series means i'm going to multiply a positive times a negative at the end.

02:15 So for example, i would have negative 3 times negative 3, which is positive 9 times negative 3 will give me negative 27.

02:22 And then negative 27 times negative 3 would give me positive 81.

02:27 So basically what i'm saying is that it's negative because the exponent is odd.

02:33 So we'll have negative 243 here.

02:40 Minus three times three is nine minus three plus five.

02:47 And we just want to crunch those numbers.

02:50 So we're going to say negative 243 minus nine.

03:03 And i encourage you to do this as well.

03:04 Negative 243 minus nine minus three plus five gives us negative 250.

03:15 All right.

03:16 So let's do the same thing with the negative three.

03:19 So this is going to be a little different because our x here is negative.