Question

f(x) = kx + x^2 – x^3

The graph of y=f(x) has no turning points. Find the
range of
values of k.

Answer 1

Answer:

Solution

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Given(3t−2)/4−(2t+3)/3=2/3−t

Now by taking L.C.M for 4 and 3 is 12

(3(3t−2)−4(2t+3))/12=2/3−t

By transposing the above equation we can write as

(9t−6−8t−12)/12=2/3−t

t+12t=8+18

Again by transposing

t = 26/13

t = 2

Answer 2

Answer:

The range of k is ( -∞ , $\frac{-1}{3}$ )

Step-by-step explanation:

For the function to have no turning point, the slope of the function must not become zero at any point in domain.

                                  $f(x)=kx+x^2-x^3$

Derivative of f(x) will be given as,

                                  $f'(x)=k+2x-3x^2$

Now the derivative must not have any roots so solving and putting discriminant less than zero.

                                  $3x^2-2x-k=0\\D=4-4(3)(-k)\\D < 0\\4+12k < 0\\k < \frac{-1}{3}$

Hence the range of k is ( -∞ , $\frac{-1}{3}$ )