Find the critical number of the function. 1. f(x) = 5x^2 + 7x

To find the critical values  of the function f(x) =
5x^2+7x.

Solution:

Critical
points of the functions are those points where the tagents are to the curve is either
parallel to x axis or vertical to the x axis, or  where the curve crosses the
axis.

The tangents are parallel to x axis  when f'(x) =
0.

Critical values of a function are the values of the
function at critical points.

f(x) =
5x^2+7x.

The curve crosses x axis when  f(x) =
0.

So f(x) = 0 gives 5x^2+7x = 0. Or x(5x+7) = 0 .
Therefore x= 0 or 5x+7 = 0 gives x = -7/10. Therefore x = 0 is a critical point, and x =
-7/5 is also a critical point.f(0) = 0 and f(-7/5) = 0 are  the critical
values.

Now consider for the critical points  when f'(x) =
0. Or when (5x^2+7x)' = 0. Or  when (5*2x+7) = 0. Or 10x = -7. So x = -7/10. Therefore x
= -7/10 is a critical point where  dy/dx = 0 . So x= -7/10 is a  critical point where
the tangent to the curve f(x) = 5x^2+7x is || to x axis. The critical value
corresponding to the critacal point x = -7/10 is f(-7/10) = 5(-7/10)^2 +7(-7/10) =
-2.45.