QuestionConsider the following. (If an answer does not exist, enter DNE.) $$f(x)=x^{3}-3 x^{2}-7 x+5$$
Find the interval(s) on which $f$ is concave up. (Enter your answer using interval notation.) $\square$ Find the interval(s) on which $f$ is concave down. (Enter your answer using interval notation.) $\square$ Find the inflection point of $f$. $\square$ $$(x, y)=(\square)$$ Need Help? Read It

Studdy Solution
Find the inflection point(s), where the concavity changes.
Since $x = 1$ is where $f''(x)$ changes sign, it is an inflection point.
Calculate $f(1)$:
$$f(1) = (1)^3 - 3(1)^2 - 7(1) + 5$$
$$f(1) = 1 - 3 - 7 + 5 = -4$$
The inflection point is $(1, -4)$.
The intervals and inflection point are: - Concave up: $(1, \infty)$ - Concave down: $(-\infty, 1)$ - Inflection point: $(1, -4)$