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

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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.

The critical number of the function is the value of x that cancels the first derivative of the function.

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

f'(x) = 5*2x^1 + 7

f'(x) = 10x + 7

Now, we'll calculate the roots of the equation f'(x) = 0.

10x + 7 = 0

We'll subtract 7 both sides:

10x = -7

x = -7/10

**The critical value for f(x) is x = -7/10.**

Now, we can compute the local extreme of the given function, substituting x by the critical value.

f(-7/10) = 5(-7/10)^2 + 7(-7/10)

f(-7/10) = 245/100 - 49/10

f(-7/10) = (245-490)/100

**f(-7/10) = -245/100**

**The minimum point of the function is the vertex of the parabola and it has the coordinates: (-7/10 ; -245/100).**