Question

The derivative of f is given f'(x) = e^x(-x^3+3x)-3 for [0,5]. What value of f(x) is an absolute minimum?

Answer 1

\[f'(x) = e^x (-x^3+3x)-3 \]for \[0 \le x \le 5\]

Answer 2

I used a graphing calculator to graph the derivative and found the absolute minimum value to be 5, but how would you do this without a graphing calculator?

Answer 3

at a maxima or minima, you have
\[f'(x)=0\]
solve this equation for "x" in the given interval

Answer 4

it's kinda hard to solve for x...?

Answer 5

ok. In the given interval,
1) \(e^x\) is continuously increasing an positive
2) \[-x^3+3x=x(3-x^2)=\left\{\begin{matrix}>0&\forall&0\le x<\sqrt{3}\\=0&&x=\sqrt{3}\\<0&\forall&\sqrt{3}<x\le5\end{matrix}\right.\]

Answer 6

what's the upside down A stand for?

Answer 7

hence, we notice that the derivative was -ve -> f(x) was decreasing
then becomes +ve -> f(x) increases
then -ve again -> f(x) decreases

Answer 8

so the chances of having the absolute minima would be at x=5

Answer 9

to find the minimum value, we need some information about the function

Answer 10

upside-down A means -> for all values of

Answer 11

oh okay i think i'm understanding your logic there.

Answer 12

Remember, we know that local extrema occur where the derivative is 0 or does not exist, and these are the critical values.

So find the critical values, but in this case we know that the graph will be differentiable everywhere so we only find where it is equal to 0. By finding where it is 0 will give you the critical values. Then find the second derivative, f''(x) and use it find whether function is concave up or concave down for the critical values you determined. If f''(x) > 0 then the function is concave up, which means the point is a minimum, otherwise if f''(x) < 0 then the function is concave down which means that there is a maximum. Can you follow this and find the where there is a minimum?

@jennychan12

Answer 13

Btw, by absolute minimum, do you mean the minimum absolute value? @jennychan12

Answer 14

i think it was just looking for the x-coordinate.

Answer 15

Well if it is just in terms of local minimum, then do what I mentioned there. But I believe you will need a graphing calculator to find where f'(x) = 0

Answer 16

yeah that's the problem. and it's in the no calculator portion of the "practice test"

Answer 17

Ok so I got it. We integrate the function like this:\[\int\limits_{}^{}[e^x(-x^3+3x)-3]dx=\int\limits_{}^{}e^x(-x^3+3x)dx-\int\limits_{}^{}(3)dx\]We can use tabular integration to integrate the first integral and we can quite easily integrate the second integral. I can't exactly show my work here for tabular integration but I'm assuming you know how to do it (it's extremely simple and efficient especially for indefinitely integrals like these containing e^x). The final indefinite integral through tabular integration results in:\[f(x)=-e^x(x^3-3x^2+3x-3)-3x+C\]We here notice that -e^x is negative everywhere and is decreasing as x approaches infinity. So basically, in order to minimise the value of f(x), the polynomial in the bracket must be the largest value possible in [0,5] for the entire thing to be the smallest possible value (minimum) for f(x) in [0,5]. We also know that the derivative of this polynomial is:\[\frac{ d }{ dx }x^3-3x^2+3x-3=3x^2-3x+3=3(x^2-x+1)\]This derivative is positve for [0,5] which means that the original polynomial is increasing in [0,5] which implies that the maximum value of the polynomial occurs at x = 5, therefore the minimum value for f(x) occurs at x = 5 because this is where the polynomial in brackets is the maximum value in the given interval which makes the whole f(x) value smallest possible in the interval [0,5].

Therefore the minimum for f(x) in the interval [0, 5] occurs at (5, f(5)).

Note: We don't know f(5) since f(x) is an indefinite integral and we are not aware of the constant at the end which could affect the value of f(5). If we were given an initial value, then we could've figured out the value of the constant.

Answer 18

@jennychan12

Answer 19

-_- wow.
okay, thanks.