# What are the extrema of f(x)=-3x^2+30x-74 on [-oo,oo]?

Oct 11, 2017

Let's see.

#### Explanation:

Let the function given be $y$ such that $\rightarrow$ for any value of $x$ in the given range.

$y = f \left(x\right) = - 3 {x}^{2} + 30 x - 74$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - 6 x + 30$

$\therefore \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - 6$

Now, since the second order derivative of the function is negative, the value of $f \left(x\right)$ will be maximum.

Hence, point of maxima or extrema can only be obtained.

Now, whether for maxima or minima,

$\frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\therefore - 6 x + 30 = 0$

$\therefore 6 x = 30$

$\therefore x = 5$

Therefore, the point of maxima is $5$. (Answer).

So, the maximum value or the extreme value of $f \left(x\right)$ is $f \left(5\right)$.

$\therefore f \left(5\right) = - 3. {\left(5\right)}^{2} + 30.5 - 74$

$\therefore f \left(5\right) = - 75 + 150 - 74$

$\therefore f \left(5\right) = 150 - 149$

$\therefore f \left(5\right) = 1$.

Hope it Helps:)