What is the solution to (x - 5)(x + 2)(x - 7) <0?

For x < -2  f(x) < 0.

For   5 < x < 7 , f(x) < 0.

In interval notation it is:

(∞, -2) ∪ ( 5, 7).

Step-by-step explanation:

First find the critical values of x:

f(x) = (x - 5)(x + 2)(x - 7) = 0

x = 5, -2  and 7.

(x - 5)(x + 2)(x - 7) < 0

Consider x = -2:

When x < -2 the sign of the expression is - * - * - = -  

and it is 0 for x = -2

So for x < -2,  f(x) < 0.

For:

-2  < x < 5  the sign of f(x) is - * + * -  = +

So for  -2  < x < 5,  f(x) > 0

For:

5 < x < 7  the sign of f(x) is + * + * - = -

So for 5 < x < 7 , f(x) < 0.

For x > 7,   f(x) > 0.

answer: on graphs, the open and closed circles, or vertical asymptotes drawn as dashed lines us identify discontinuities.

as before, graphs and tables allow us to estimate at best.

when working with formulas, getting zero in the denominator indicates a point of discontinuity.

when working with piecewise-defined functions, check for discontinuities at the transition points where one piece ends and the next begins.

step-by-step explanation:

i suppose it should be (y=2x-5)

step-by-step explanation: