Question

Let f(x) =x^2+10x+29

What is the minimum value of the function?

____

Answer 1

The answer for it is 4 at x=-5. You need to take a derivative of you function and set f'(x)=0, then you find critical points. Set those x values into f(x) and take the lowest value.

Answer 2

Hence, the minimum value of the function is 4.

Explanation:

We re given the function
`f(x)=x^2+10x+29`

Differentiating with respect to x, we get,

`f'(x)=2x+10`

Equating f'(x) to 0, we have,

`f'(x)=0`

i.e.
`2x+10=0`

i.e.
`2x=-10`

i.e.
`x=-5`

Now, differentiating f'(x) with respect to x gives us,

`f''(x)=2>0`

Thus, the function f(x) have minimum at x= -5 and the minimum value is,

`f(-5)=(-5)^2+10* (-5)+29`

i.e.
`f(-5)=25-50+29`

i.e.
`f(-5)=54-50`

i.e. f(-5) = 4

Hence, the minimum value of the function is 4.