how do u show a curve has no turning points? dy/dx is -7/(x-4)^2?

Answer 1

dy/dx = -7/(x-4)^2

for turning points dy/dx = 0
=> -7/(x-4)^2 = 0
not possible . if A/B = 0 => A = 0
=> no turning points

OR

if dy/dx < 0 => decreasing
-7/(x-4)^2 < 0
-7 = - no. & (x-4)^2 = +no.
-no/+no = - no

=> decreasing all the time ( except when x=4)

=> no turning points

Answer 2

At the turning point, dy/dx = 0.
When you substitute in a negative number of x e.g -1,
dy/dx = (-7) / (-1-4)^2
= -7 / 25 [negative]

When you substitute in a positive number of x e.g 1,
dy/dx = (-7) / (1-4)^2
= -7/9 [negative]

Since both are negative values, it shows that both gradients around the critical value of dy/dx=0 are downward slopes, hence there are no turning points in this curve.

From the dy/dx you can also conclude the same thing because the denominator is squared, hence no matter what x you insert into the dy/dx the denominator will ALWAYS be positive. Since the numerator is a constant (-7), hence the dy/dx will always be negative, thus no turning point.

** If a curve has turning points, the substitution of a slightly negative value and a slightly positive value into the dy/dx should give you values of opp signs.
E.g
negative x gives negative dy/dx
positive x gives positive dy/dx

or
negative x gives positive dy/dx
positive x gives negative dy/dx

This will give you a curve which has a turning point.

Answer 3

Recall that:

At the turning points dy/dx = 0 and it changes sign either from positive to negative ( a maxima point) or from negative to positive ( a minima point).

In your case: dy/dx = - 7/(x - 4)^2, which will always remain negative due to - 7. The quantity (x - 4)^2, being a square, will always be positive for all real values of x. It can never be zero so the curve will never have any turning point.

Answer 4

dy / dx = 0 for turning points
But -7 / (x - 4)² = 0 has no solution.
Therefore no turning points.

Answer 5

if dy/dx does not change sign at the critical value, where dy/dx is o or undfined (when x=4 in this case), then the curve has no turning points there

Answer 6

unless im being too simplistic, that is a quadratic graph and it does have a turning point,
its of the form y=a(x-h)^2+k where t.p.=(h,k)
and can be otherwise written as:

(-1/7)(x - 4)^2+0

therefore the turning point is (4,0) as k=0

the graph is negative and stretched to the x axis, sorry its a bit hard to show through text

anyway hope this helps, im going to look like an idiot if its some really hard problem and ive interpreted it wrong.