3 years ago

Evaluate 1x^3 –2x^2 – 4x- 6 at x = 4

Mathematics

1 answer:

Whitepunk [10]3 years ago

4 0

Answer:

Step-by-step explanation:

1x^3-2x^2-4x-6

x^3-2x^2-4x-6

(4)^3-2(4)^2-4(4)-6

64-2(16)-16-6

64-32-16-6

32-16-6

16-6

So when x=4, the output will be 10

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Assume that f is continuous on [-4,4] and differentiable on (-4,4). The table gives some values of f'(x) x: -4, -3, -2, -1, 0, 1

kondaur [170]

$f$ will be increasing on the intervals where $f'(x)>0$ and decreasing wherever $f'(x)$ . Local extrema occur when $f'(x)=0$ and the sign of $f'(x)$ changes to either side of that point.

$f'(x)$ is positive when $x$ is between -4 and some number between -2 and -1, and also 2 (exclusive) and 4, so you can estimate that $f(x)$ is increasing on the intervals [-4, -2] and (2, 4].

$f'(x)$ is negative when $x$ is between some number between -2 and -1, up to some number less than 2. So $f(x)$ is decreasing on the interval [-1, 1].

You then have two possible cases for extrema occurring. The sign of $f'(x)$ changes for some $x$ between -2 and -1, and again to either side of $x=2$ .