Find the minimum value if f(x) = xe^x over [-2,0]

Mathematics

1 answer:

SashulF [63]1 year ago

6 0

Given function:

$f(x)=xe^x$

The minimum value of the function can be found by setting the first derivative of the function to zero.

$f^{\prime}(x)=xe^x+e^x$ $\begin{gathered} xe^x+e^x\text{ = 0} \\ e^x(x\text{ + 1) = 0} \end{gathered}$

Solving for x:

$\begin{gathered} x\text{ + 1 = 0} \\ x\text{ = -1} \end{gathered}$ $\begin{gathered} e^x\text{ = 0} \\ \text{Does not exist} \end{gathered}$

Substituting the value of x into the original function:

$\begin{gathered} f(x=1)=-1\times e^{^{-1}}_{} \\ =\text{ -}0.368 \end{gathered}$

Hence, the minimum value in the given range is (-1, -0.368)

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Find the quadratic function y=f(x) whose graph has a vertex (−3,4) and passes through the point (−7,0). Write the function

olganol [36]

Answer:

Step-by-step explanation:

This is a parabola since a quadratic is a parabola. The standard form for a parabola is y = ax² + bx + c

but before we do that, we will use the vertex form, since it will make our work easier at the beginning.

First and foremost, when we plot the vertex and the given point, the vertex is higher up than is the point; that means that this parabola opens upside down, and its vertex form will be

y=-|a|(x - h)² + k

The absolute value is out in front of the a, so we know that the value of a is positive, but the quadratic itself is negative (upside down) and we will find that math takes care of that negative that needs to be out front. So we need to solve for a by filling in the x, y, h, and k values from the point and the vertex: x = -7, y = 0, h = -3, k = 4

0 = a(-7 - (-3))² + 4 and

0 = a(-7 + 3)² + 4 and

0 = a(-4)² + 4 and

0 = a(16) + 4 and

0 = 16a + 4 and

-4 = 16a so

$a=-\frac{1}{4}$