Question

For the given curve: y = 5x – 2x3, when x increases at the rate of 2 units/sec, then how fast is the slope of curve changes when x = 3?

Show that the function f(x) = tan x – 4x is strictly decreasing on [-π/3, π/3]​

Answer 1

Here is your answer:-

1. Given that, y = 5x – 2x3

Then, the slope of the curve, dy/dx = 5-6x2

⇒d/dt [dy/dx]= -12x. dx/dt

= -12(3)(2)

= -72 units per second

Hence, the slope of the curve is decreasing at the rate of 72 units per second when x is increasing at the rate of 2 units per second.

2. Given that, f(x) = tan x – 4x

Then, the differentiation of the function is given by:

Then, the differentiation of the function is given by:f’(x)= sec2x – 4

When -π/3 <x π/3, 1<sec x <2

Then, 1<sec2x <4

Hence, it becomes -3 < (sec2x-4)<0

Hence, for -π/3 <x π/3, f’(x)<0

Therefore, the function “f” is strictly decreasing on [-π/3, π/3].

Hope this helps..❤️

Answer 2

Answer:

1. Given that, y = 5x – 2x3

Then, the slope of the curve, dy/dx = 5-6x2

⇒d/dt [dy/dx]= -12x. dx/dt

= -12(3)(2)

= -72 units per second

Hence, the slope of the curve is decreasing at the rate of 72 units per second when x is increasing at the rate of 2 units per second.

2. Given that, f(x) = tan x – 4x

Then, the differentiation of the function is given by:

Then, the differentiation of the function is given by:f’(x)= sec2x – 4

When -π/3 <x π/3, 1<sec x <2

Then, 1<sec2x <4

Hence, it becomes -3 < (sec2x-4)<0

Hence, for -π/3 <x π/3, f’(x)<0

Therefore, the function “f” is strictly decreasing on [-π/3, π/3].