Question

Select the correct answer from each drop-down menu. Consider functions h and k. h(t) = 5x^2 - 1k(u) = √5x+1 For I > 0, the value of h(k(x)) is ____the value of k(h(x))options-equal tonot equal toFor I > 0, functions h and k ____ inverse functionsoptions- areare not

Answer 1

SOLUTION

Given the question in the image, the following are the solution steps to answer the question.

STEP 1: Write the given functions

$\begin{gathered} h(x)=5x^2-1 \\ k(x)=\sqrt{5x+1} \end{gathered}$

STEP 2: Find h(k(x))

$\begin{gathered} We\text{ insert k\lparen x\rparen into h\lparen x\rparen as seen below} \\ x\Rightarrow\sqrt{5x+1} \\ h(k(x))=5(\sqrt{5x+1})^2-1 \\ =5(5x+1)-1 \\ =25x+5-1=25x+4 \end{gathered}$

STEP 3: Find k(h(x))

$\begin{gathered} We\text{ insert h\lparen x\rparen into k\lparen x\rparen as seen below:} \\ x\Rightarrow5x^2-1 \\ k(h(x))=\sqrt{5(5x^2-1)+1} \\ =\sqrt{25x^2-5+1}=\sqrt{25x^2-4} \\ \\ k(h(x))=\sqrt{25x^2-4} \end{gathered}$

It can be seen from above that the result for:

$h(k(x))\ne k(h(x))$

Therefore:

The value of h(k(x)) is not equal to the value of k(h(x))

Since the h of k is not equal to k of h, therefore,

h and k are no inverse function.