If the interval in which the function $f(x) = 4x^3 - 6x^2 - 72x + 30$ is strictly decreasing, is (a,b) then a+b is equal to

To find where the function $f(x) = 4x^3 - 6x^2 - 72x + 30$ is strictly decreasing, the derivative must be negative.

$f'(x) = \frac{d}{dx}(4x^3 - 6x^2 - 72x + 30)$

$f'(x) = 12x^2 - 12x - 72$

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For strictly decreasing, $f'(x) < 0$:

$12x^2 - 12x - 72 < 0$

Dividing by 12:

$x^2 - x - 6 < 0$

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Factoring the quadratic:

$x^2 - x - 6 = (x - 3)(x + 2)$

The inequality becomes:

$(x - 3)(x + 2) < 0$

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The critical points are $x = -2$ and $x = 3$.

A product is negative when one factor is positive and the other is negative.

For $x < -2$: both factors are negative, product is positive

For $-2 < x < 3$: $(x + 2) > 0$ and $(x - 3) < 0$, product is negative

For $x > 3$: both factors are positive, product is positive

The inequality $(x - 3)(x + 2) < 0$ is satisfied when $-2 < x < 3$

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The strictly decreasing interval is $(a, b) = (-2, 3)$

Therefore:

$a = -2$

$b = 3$

$a + b = -2 + 3 = 1$