Question

1. Using the Intermediate Value Theorem, verify $f(x) = 1$ at least once on the interval $[-1, 2]$. $f(x) = x^2 + 3x - 7$ 2. $\lim_{x \to \infty} \frac{5x^3 - 7x}{8x^4 - x^2 + 11} = $ 3. $\lim_{x \to \infty} \frac{8 - 7x^4}{9x^4 + 2x^3 - 3x^2 + 5} = $ 4. Find the constants $a$ and $b$ so that $f(x)$ will be a continuous function. $f(x) = \begin{cases} x^2 + b & ; x < -1 \ ax - 2 & ; -1 \le x < 3 \ 7 - 2x^2 & ; x \ge 3 \end{cases}$ 5. State whether the graph below, $f(x)$, is continuous. If it is not continuous, write the reason using one of the three conditions from the definition of continuity.

          1.
Using the Intermediate Value Theorem, verify $f(x) = 1$ at least once on the interval $[-1, 2]$.
$f(x) = x^2 + 3x - 7$
2.
$\lim_{x \to \infty} \frac{5x^3 - 7x}{8x^4 - x^2 + 11} = $
3.
$\lim_{x \to \infty} \frac{8 - 7x^4}{9x^4 + 2x^3 - 3x^2 + 5} = $
4.
Find the constants $a$ and $b$ so that $f(x)$ will be a continuous function.
$f(x) = \begin{cases} x^2 + b & ; x < -1 \ ax - 2 & ; -1 \le x < 3 \ 7 - 2x^2 & ; x \ge 3 \end{cases}$
5.
State whether the graph below, $f(x)$, is continuous. If it is not continuous, write the reason using
one of the three conditions from the definition of continuity.

1.
Using the Intermediate Value Theorem, verify f(x) = 1 at least once on the interval [-1, 2].
f(x) = x^2 + 3x - 7
2.
limx →∞(5x^3 - 7x)/(8x^4 - x^2 + 11) =
3.
limx →∞(8 - 7x^4)/(9x^4 + 2x^3 - 3x^2 + 5) =
4.
Find the constants a and b so that f(x) will be a continuous function.
f(x) = x^2 + b ; x < -1 ax - 2 ; -1 ≤ x < 3 7 - 2x^2 ; x ≥ 3
5.
State whether the graph below, f(x), is continuous. If it is not continuous, write the reason using
one of the three conditions from the definition of continuity.

Added by Christopher L.

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