Krish Schmitt

2022-10-02

Given is

$$x=3\phantom{\rule{0ex}{0ex}}h=2\phantom{\rule{0ex}{0ex}}f(x)={x}^{2}+2x+10$$

what is the solution of $f(x+h)$

Is that correct?

$$f(x)={x}^{2}+2x+10=(9+4)+2(3+2)+10=13+10+10=33$$

$$x=3\phantom{\rule{0ex}{0ex}}h=2\phantom{\rule{0ex}{0ex}}f(x)={x}^{2}+2x+10$$

what is the solution of $f(x+h)$

Is that correct?

$$f(x)={x}^{2}+2x+10=(9+4)+2(3+2)+10=13+10+10=33$$

pereishen9g

Beginner2022-10-03Added 7 answers

In the formula

$$f(x)={x}^{2}+2x+10$$

think of the "x" as a box you get to fill in, that is,

$$f(\u25fb)=(\u25fb{)}^{2}+2(\u25fb)+10$$

So you can put an x, or x+h, or anything you want, into the box. In particular, putting x+h into the box gives you

$$f(x+h)=(x+h{)}^{2}+2(x+h)+10$$

Now since x=3 and h=2, you know $\overline{){\displaystyle x+h}}=3+2=\overline{){\displaystyle 5}}$

$$f(x)={x}^{2}+2x+10$$

think of the "x" as a box you get to fill in, that is,

$$f(\u25fb)=(\u25fb{)}^{2}+2(\u25fb)+10$$

So you can put an x, or x+h, or anything you want, into the box. In particular, putting x+h into the box gives you

$$f(x+h)=(x+h{)}^{2}+2(x+h)+10$$

Now since x=3 and h=2, you know $\overline{){\displaystyle x+h}}=3+2=\overline{){\displaystyle 5}}$

Riya Andrews

Beginner2022-10-04Added 4 answers

Your basic idea is correct, but you are asked for $f(x+h)$ so that is what you should have on the left side, not $f(x)$. Then it might help to write out the function in terms of the variables first. The first term in the expression is not correct.

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