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allsm [11]
3 years ago
8

Find the maximum value or minimum value for the function f(x) = 0.15(x + 1)² - 3.

Mathematics
1 answer:
il63 [147K]3 years ago
7 0

Answer:

The minimum value for  f(x) = 0.15(x + 1)^2 - 3 is -3.

Step-by-step explanation:

Given function is f(x) = 0.15(x + 1)^2 - 3

We need to find the maximum value or the minimum value for the function.

Now, differentiate f(x) = 0.15(x + 1)^2 - 3  w.r.t x.

f'(x) =\frac{d}{dx}(0.15(x + 1)^2 - 3)\\f'(x)=\frac{d}{dx}(0.15(x+1)^2-\frac{d}{dx}(3)\\

f'(x)=2\times 0.15(x+1)\frac{d}{dx}(x+1)-0\\f'(x)=0.3(x+1)(1)\\f'(x)=0.3(x+1)

Now, we will equate f'(x)=0 to find critical point.

0.3(x+1)=0\\x=-1

Plug this critical point in to the function f(x) = 0.15(x + 1)^2 - 3  we get,

f(-1) = 0.15(-1 + 1)^2 - 3\\f(-1)=-3

Also, f''(x)=0.3 which is positive, We have minimum value.

So, the minimum value for  f(x) = 0.15(x + 1)^2 - 3 is -3.

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