In Exercises 27–38, evaluate each function at the given values of...

Question

In Exercises 27–38, evaluate each function at the given values of the independent variable and simplify. h(x) = x^4 - x²+1

c. h (-x)

Accepted Answer

Welcome back. I'm so glad you're here. We have this function H. Of X equals X to the fourth plus five X squared minus 11. And we're to evaluate for H of negative three X. What does that mean? It means that anywhere we have an X. In our original function we're going to substitute a negative three X. So there are two places we're going to do that. The first one right off the bat instead of X to the fourth. It's now going to be negative three X to the fourth plus five times no longer X squared but now negative three X squared And bring down our -11. Now we want to start off by doing our exponents so negative three X to the fourth. Well negative three to the fourth is 81 X to the fourth is X to the fourth plus. Now we're just going to bring down that five because we have to do exponents first. So negative three X squared, negative three squared is nine and X squared is X squared and bring down our minus 11. And now all we have to do is multiply that five times the nine. So this will be 81 X to the fourth plus five times nine is 45 X squared minus 11 and none of those are like terms. So that's as simplified as that one's going to get we look at our answer choices and that matches with answer choice C. Well done. We'll catch you in the next one

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In Exercises 27–38, evaluate each function at the given values of the independent variable and simplify. h(x) = x^4 - x²+1 c. h (-x)