See below for the changes when the exponential function is transformed
<h3>How to determine the effect of a</h3>
The exponential functions are given as:




An exponential function of the above form is represented as:

See attachment for the graph of the four functions.
<u>When a is large</u>
This is represented by 
In this case, the curve of the base form
is vertically stretched and it moves closer to the y-axis
<u>When a is small</u>
This is represented by 
In this case, the curve of the base form
is vertically stretched and it moves away from the x-axis
<u>When a is negative</u>
This is represented by 
In this case, the curve of the base form
is vertically stretched and is reflected across the y-axis.
Read more about function transformation at:
brainly.com/question/26896273
#SPJ1
So the -4x is the M and then 6 is the B. So m is the slope and B is the y-intercept. The 6 goes on the y axis and the -4 will be how the line looks. It’ll be rise (-4) over run (1).
These can be some values of equation
6x+ 12=30
This question might not get the right answer but this how you set equations.
If f(x) has an inverse on [a, b], then integrating by parts (take u = f(x) and dv = dx), we can show

Let
. Compute the inverse:
![f\left(f^{-1}(x)\right) = \sqrt{1 + f^{-1}(x)^3} = x \implies f^{-1}(x) = \sqrt[3]{x^2-1}](https://tex.z-dn.net/?f=f%5Cleft%28f%5E%7B-1%7D%28x%29%5Cright%29%20%3D%20%5Csqrt%7B1%20%2B%20f%5E%7B-1%7D%28x%29%5E3%7D%20%3D%20x%20%5Cimplies%20f%5E%7B-1%7D%28x%29%20%3D%20%5Csqrt%5B3%5D%7Bx%5E2-1%7D)
and we immediately notice that
.
So, we can write the given integral as

Splitting up terms and replacing
in the first integral, we get
