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Chapter 4: Problem 61

Determine the intervals on which the following functions are concave up or concave down. Identify any inflection points. $$f(x)=e^{x}(x-3)$$

Short Answer

Expert verified
Question: Determine the intervals of concavity and the inflection point for the function $$f(x) = e^x(x-3).$$ Answer: The intervals of concavity for the function are concave up on $$(1, \infty)$$ and concave down on $$(-\infty, 1).$$ The inflection point is at $$(1, -2e).$$

Step by step solution

01

Find the first derivative of the function f(x)

To find the first derivative of the function $$f(x) = e^x(x-3),$$ use the product rule: $$(uv)' = u'v + uv'.$$ Here, let $$u = e^x$$ and $$v = x-3.$$ The derivatives of u and v are: $$u' = \frac{d}{dx}(e^x) = e^x$$ $$v' = \frac{d}{dx}(x-3) = 1.$$ Now apply the product rule: $$f'(x) = u'v + uv' = e^x(x-3) + e^x(1) = e^x(x-2).$$
02

Find the second derivative of the function f(x)

Next, find the second derivative of the function by differentiating the first derivative: $$f''(x) = \frac{d^2}{dx^2}(e^x(x-2)).$$ Use the product rule again, but this time use $$u = e^x$$ and $$v = x-2.$$ We already found $$u' = e^x$$ in Step 1. Differentiate v to find $$v' = \frac{d}{dx}(x-2) = 1.$$ Then apply the product rule: $$f''(x) = u'v + uv' = e^x(x-2) + e^x(1) = e^x(x-1).$$
03

Determine the sign of f''(x)

Now, analyze the sign of the second derivative $$f''(x) = e^x(x-1).$$ Since $$e^x > 0$$ for all x values, the sign of the second derivative is determined by the factor $$(x-1).$$ So, $$f''(x) > 0$$ for $$x>1$$ and $$f''(x) < 0$$ for $$x<1.$$
04

Determine the intervals of concavity and inflection points

From Step 3, we can see that the function is concave up when $$f''(x) > 0$$, that is $$x>1.$$ The function is concave down when $$f''(x) < 0$$, that is $$x<1.$$ Therefore, the intervals of concavity are as follows: - Concave up: $$(1, \infty)$$ - Concave down: $$(-\infty, 1)$$ The function has an inflection point at the value of x where the second derivative changes its sign, which is $$x=1.$$ To find the corresponding y-value, plug this value of x into the original function: $$f(1) = e^1(1-3) = -2e.$$ Thus, the inflection point is at $$(1, -2e).$$

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