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# Find the absolute maximum and minimum values of f(x)=4x+4x^2, if any, over the interval (−∞,+∞).?

### 3 Answers

- Wayne DeguManLv 72 months agoFavorite Answer
f '(x) = 4 + 8x

Stationary points occur when f '(x) = 0

i.e. 4 + 8x = 0

so, x = -1/2

f''(x) = 8...i.e. minimum as f''(x) > 0

f(-1/2) = 4(-1/2) + 4(-1/2)² => -1

Hence, miimum at (-1/2, -1)

As x --> ∞, f(x) --> ∞

As x --> -∞, f(x) --> ∞

:)>

- lenpol7Lv 72 months ago
Differentiate and equate to zero .

f(x) = 4x + 4x^2

f'(x) = 4 + 8x = 0

8x = -4

x = -1/2

When x = -1/2

f(-1/2) = 4(-1/2) + 4(-1/2)^2 = -2 + 1 = -1

Hence the turning point is ( -1/2 , -1)

We differentiate again. If the result is positive(+)/Negative(-) then it is minimum/Maximum.

Hence f'(x) = 4 + 8x

f''(x) = (+)8 ; result is positive so it is a minimum at the turning point.

NB When x = +/−∞ Then y = +∞ at a maximum.

- MorningfoxLv 72 months ago
Hint: for x very very large and positive, x^2 is super big.

Hint: for x very very large and negative, x^2 is super big.