Question

The nth derivative of the function t(x) where t(x)=5x^6 - 9x^5 +3x^3 - 0.5 is a cubic function.
(a)State the value of n.
(b)Find the ratio of coefficient in x^3 to the coefficient in x^2 of the cubic function giving your answer in the form 1:k(1 is to k)where k is a fraction in its simplest form.

Answer 1

Answer on Question #53276 – Math – Calculus

Question

The $n$th derivative of the function $t(x)$ where $t(x) = 5x^6 - 9x^5 + 3x^3 - 0.5$ is a cubic function.

(a) State the value of $n$.

(b) Find the ratio of coefficient in $x^3$ to the coefficient in $x^2$ of the cubic function giving your answer in the form $1:k(1 \text{ is to } k)$ where $k$ is a fraction in its simplest form.

Solution

(a) Let us take derivatives of $t(x)$ until we get a cubic function.

$$\frac{d}{dx} t(x) = 30x^5 - 45x^4 + 9x^2$$ $$\frac{d^2}{dx^2} t(x) = \frac{d}{dx} \left( \frac{d}{dx} t(x) \right) = 150x^4 - 180x^3 + 18x$$ $$\frac{d^3}{dx^3} t(x) = \frac{d}{dx} \left( \frac{d^2}{dx^2} t(x) \right) = 600x^3 - 540x^2 + 18$$

$\frac{d^3}{dx^3} t(x)$ is cubic function, thus, $3^{rd}$ derivative of the function $t(x)$ is a cubic function, hence $n = 3$.

(b) The ratio of coefficient in $x^3$ to the coefficient in $x^2$:

$$\frac{600}{-540} = -\frac{60}{54} = -\frac{20}{18} = -\frac{10}{9}$$

Answer:

(a) 3

(b) $-\frac{10}{9}$

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