I need HELP
4.The derivative of a function f is given by f'(x)=(-2x-2)e^x, and f(0) = 3.
A. The function f has a critical point at x = -1. At this point, does f have a relative minimum, a relative maximum, or neither? Justify your answer.
B. On what intervals, if any, is the graph of f both increasing and concave down? Explain your reasoning.
C. Find the value of f(-1).
f'(x) = -2(x+1)e^x
f"(x) = -2(x+2)e^x
(A) since f"(-1) < 0, f is concave downward -- has a maximum there
(B) increasing: x > -1
concave down: x < -2
so, both: never
(C) f(x) = -2xe^x + C
since f(0) = 3, C = -3 and
f(x) = -2xe^x + 3
f(-1) = 2/e + 3
sure -- because that's what it is
f'(x) = (-2x-2)e^x = -2xe^x - 2e^x
f"(x) = -2e^x - 2xe^x - 2e^x = -2xe^x - 4e^x = -2(x+2)e^x
or are you asking why I used f" as the label?
Also because that's what's often used.
Then nth derivative for n > 2 is usually f(n)(x)
Hey thank you so much for the response. So is there a reason why you labeled the -2(x+2)e^x equation as a second derivative with f"
Oh nevermind, that makes more sense, thank you though
To answer these questions, we need to understand the properties of derivatives and their relation to the original function.
A. To determine the nature of the critical point at x = -1, we can use the Second Derivative Test. The Second Derivative Test states that:
- If f''(x) > 0, then f has a relative minimum at x.
- If f''(x) < 0, then f has a relative maximum at x.
- If f''(x) = 0, the test is inconclusive.
To apply the Second Derivative Test, we need to find the second derivative of f(x). Let's start by finding the first derivative of f(x) and then the second derivative.
Given: f'(x) = (-2x-2)e^x
To find f''(x), we differentiate f'(x) with respect to x:
f''(x) = d/dx[(-2x-2)e^x]
= (-2e^x) - 2e^x
= -4e^x
Now we have the second derivative f''(x) = -4e^x.
Since f''(-1) = -4e^(-1) < 0, it means that the second derivative at x = -1 is negative. According to the Second Derivative Test, a negative second derivative indicates a relative maximum.
Therefore, at x = -1, the function f has a relative maximum.
B. To determine the intervals where the graph of f is both increasing and concave down, we need to analyze the signs of the first and second derivatives.
- If f'(x) > 0 and f''(x) < 0, the graph of f is increasing and concave down.
- If f'(x) > 0 and f''(x) > 0, the graph of f is increasing and concave up.
- If f'(x) < 0 and f''(x) > 0, the graph of f is decreasing and concave up.
- If f'(x) < 0 and f''(x) < 0, the graph of f is decreasing and concave down.
We know that f'(x) = (-2x-2)e^x and f''(x) = -4e^x.
To find the intervals, we need to determine where the first derivative is positive and the second derivative is negative.
When f'(x) = (-2x-2)e^x > 0, it means that (-2x-2) and e^x have the same sign. Since e^x is always positive, (-2x-2) must be positive. Dividing by -2, we get x+1 < 0, which corresponds to x < -1.
So, f'(x) > 0 for x < -1.
Now let's determine where f''(x) = -4e^x < 0:
-4e^x < 0
e^x > 0
Since e^x is always positive, f''(x) is negative for all x.
Therefore, the graph of f is both increasing and concave down for x < -1.
C. To find the value of f(-1), we can substitute x = -1 into the original function f(x).
Given: f(x) = ?
f'(-1) = (-2(-1)-2)e^(-1) = (2-2)e^(-1) = 0
Now, to find f(x), we can integrate f'(x) with respect to x:
f(x) = ∫f'(x) dx
f(x) = ∫((-2x-2)e^x) dx = (-2∫x e^x dx) - 2∫e^x dx
We can integrate each term separately:
∫x e^x dx = xe^x - ∫e^x dx = xe^x - e^x + C1
∫e^x dx = e^x + C2
Therefore,
f(x) = (-2(xe^x - e^x + C1)) - 2(e^x + C2) + C3
= -2xe^x + 2e^x + 2 - 2e^x + C1 + 2e^x + C2 + C3
= -2xe^x + 2 + C
Where C = (C1 + C2 + C3) is the constant of integration.
Since f(0) = 3, we can substitute this value into the equation to find C:
-2(0)e^0 + 2 + C = 3
C = 3 - 2
C = 1
Therefore, the value of f(-1) is given by:
f(-1) = -2(-1)e^(-1) + 2 + 1
= 2(e^(-1)) + 3
So, the value of f(-1) is 2(e^(-1)) + 3.