Let's solve the given problem step-by-step. The function is defined as $f(x) = x^2 + 1$. Now, we'll compute the values required in each part.

(a) $f(t+1)$

We substitute $t+1$ into the function: $f(t+1) = (t+1)^2 + 1$ Expand the square: $f(t+1) = t^2 + 2t + 1 + 1 = t^2 + 2t + 2$

(b) $f(t^2 + 1)$

Substitute $t^2 + 1$ into the function: $f(t^2 + 1) = (t^2 + 1)^2 + 1$ First, expand $(t^2 + 1)^2$: $(t^2 + 1)^2 = t^4 + 2t^2 + 1$ Now, add 1: $f(t^2 + 1) = t^4 + 2t^2 + 1 + 1 = t^4 + 2t^2 + 2$

(c) $f(2)$

Substitute 2 into the function: $f(2) = 2^2 + 1 = 4 + 1 = 5$

(d) $2f(t)$

First, find $f(t)$: $f(t) = t^2 + 1$ Now, multiply by 2: $2f(t) = 2(t^2 + 1) = 2t^2 + 2$

(e) $(f(t))^2 + 1$

Find $(f(t))^2$ first: $f(t) = t^2 + 1$ Square it: $(f(t))^2 = (t^2 + 1)^2 = t^4 + 2t^2 + 1$ Now, add 1: $(f(t))^2 + 1 = t^4 + 2t^2 + 1 + 1 = t^4 + 2t^2 + 2$

Summary of Results:

• (a) $f(t+1) = t^2 + 2t + 2$
• (b) $f(t^2 + 1) = t^4 + 2t^2 + 2$
• (c) $f(2) = 5$
• (d) $2f(t) = 2t^2 + 2$
• (e) $(f(t))^2 + 1 = t^4 + 2t^2 + 2$

Would you like more details or explanations on any part?

Related Questions:

1. What happens if we differentiate $f(x) = x^2 + 1$?
2. How does the function $f(x) = x^2 + 1$ change when shifted horizontally by 1?
3. What is the domain and range of the function $f(x) = x^2 + 1$?
4. How do you compute $f(f(t))$ based on $f(x) = x^2 + 1$?
5. Can we generalize $f(x) = x^2 + 1$ to a higher degree polynomial?

Tip:

When substituting expressions into a function, carefully expand the terms before simplifying to avoid common algebraic errors.