Question

Find the open interval where the function f(x)= 1/3 x^3 -3x^2+5x-7 is concave down

Answer 1

Concavity interval definitions is given as

If f"(x) > 0, then f(x) concave upwards

If f"(x) < 0, then f(x) concave downwards

`f^(\prime)^(\prime)(x)=2x-6`

equating to zero, we have

[tex]concave\text{ }downward:-\inftyso, cocave downwards becomes [tex]-\infty