Question

Let f be the function given by f(x) = (x2 - 2x - 1)e". (a) Find lim f(x) and lim (x). lim fx=(18-21 li)=2" = 0 (b) Find the intervals on which is increasing Show the analysis that leads to your answer. (c) Find the intervals on which the graph off is concave downward. Show the analysis that leads to your answer. d) Sketch the graph off.

Answer 1

To find the limits, intervals of increase and concavity, and sketch the graph of a function f(x) = (x^2 - 2x - 1)e^x.

Step-by-step explanation:

To find the limit of the function f(x), we substitute the limit value of x into the function and simplify. In this case, f(x) = (x^2 - 2x - 1)e^x. Taking the limit as x approaches a particular value will give us the function's value at that point. To find the intervals on which the function is increasing or decreasing, we can analyze its derivative. If the derivative is positive, the function is increasing. If the derivative is negative, the function is decreasing. To find the intervals on which the graph of f is concave downward, we need to analyze the second derivative. If the second derivative is negative, the graph is concave downward. Finally, we can sketch the graph of f by plotting key points and connecting them with smooth curves based on the behavior of the function and its derivatives.

Answer 2

The limit as x approaches ∞ for f(x) is ∞, while as x approaches -∞, f(x) approaches 0. f(x) is increasing for all x and is never concave downward for any interval.

(a) To find
`\(\lim_(x \to \infty) f(x)\) and \(\lim_(x \to -\infty) f(x)\)`:

`\[f(x) = (x^2 - 2x - 1)e^x\]`

As x approaches positive or negative infinity, the behavior of the exponential term
`\(e^x\)` dominates. Since
`\(e^x\)` grows without bound as x goes to infinity and approaches zero as x goes to negative infinity, the limits will be determined by the behavior of
`\(e^x\)`.

Therefore:

-
`\(\lim_(x \to \infty) f(x) = \infty\)` because
`\(e^x\)` grows without bound as x approaches infinity.

-
`\(\lim_(x \to -\infty) f(x) = 0\)` because
`\(e^x\)` approaches zero as x approaches negative infinity.

(b) To find the intervals on which f is increasing, we'll find the derivative of f(x) and then determine where the derivative is positive:

Given:
`\(f(x) = (x^2 - 2x - 1)e^x\)`

Let's find f'(x) (the derivative of f(x)) using the product rule and determine its sign:

`\[f'(x) = (2x - 2)e^x + (x^2 - 2x - 1)e^x = (x^2 + 1)e^x\]`

For f'(x) to be positive,
`\(e^x\)` must be positive (as
`\(x^2 + 1\)` is always positive). Since
`\(e^x\)` is positive for all real values of x, f'(x) will be positive for all real values of x.

Therefore, f(x) is increasing for all real values of x.

(c) To find the intervals on which the graph of f is concave downward, we'll find the second derivative of f(x) and determine where it's negative:

Given:
`\(f(x) = (x^2 - 2x - 1)e^x\)`

Let's find f''(x) (the second derivative of f(x):

`\[f'(x) = (x^2 + 1)e^x\]`

`\[f''(x) = (2x)e^x + (x^2 + 1)e^x = (x^2 + 2x + 1)e^x\]`

For f''(x) to be negative,
`\((x^2 + 2x + 1)\)` must be negative. Factoring
`\((x^2 + 2x + 1)\)` gives us
`\((x + 1)^2\),` which is always non-negative.

Therefore, f''(x) is never negative for any real value of x, which means that the graph of f is never concave downward for any interval. It is either concave upward or has points of inflection but never concave downward for any interval.

(d)

Question:

Let f be the function given by f(x) = (x² - 2x - 1)eˣ.

(a) Find
`\lim_(n \to \infty) f(x)` and
`\lim_(n \to -\infty) f(x)`.

(b) Find the intervals on which f is increasing. Show the analysis that leads to your answer.

(c) Find the intervals on which the graph off is concave downward. Show the analysis that leads to your answer.

(d) Sketch the graph of f.