Coding a plot with functions [closed] - r

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So I've probably referenced the entire internet trying to make this problem work, and haven't. However, I found stack overflow. Like I said I've been learning for not even 2 weeks yet.
So this is the problem
Let
f(x)=sqrt((x^3+3x^2+1)/(x^4+5x^3+7x+9))
(x ≥ 0)
(a) Draw a line graph of (x, f(x)) for 0 ≤ x ≤ 10 with increments of 0.01
(b) Find numerically the maximum value of f(x) and the maximizer x (report x to the
second decimal place. For instance, x = 1.23)
So I'm basically been saying x=x and y= the sqrt....., and then I write plot(x,y,type="l") and usually it just doesn't even work.
Also how do I do the increment part. I'm sorry for lack of explanation, but I have no idea what most of this means.

Answer 1

First thing to do would be to define the function:
equation <- function(x){
sqrt((x^3+3*x^2+1)/(x^4+5*x^3+7*x+9))
}
Then, define the values you want to apply the function to, and store them in vector input
input<-seq(0,10,0.01)
Apply the equation function to input, and store the values in vector results
results<-sapply(input,function)
Produce a line plot:
plot(input,results,type="l")
Print the value of x which maximises f(x)
maxx<-input[which.max(results)]
maxx

Answer 2

I would suggest a ggplot2 approach. First you have to create a random x variable and then compute y. I will add the code for that variables an the plot.
In the case of finding the maximum of f(x) you must know calculus or you can use a visual approach. Here the code:
library(ggplot2)
library(dplyr)
set.seed(123)
First we create a random variable x with the limits you mentioned:
#Data
x <- runif(100,0,10)
Now, we save the variable in a dataframe and compute y:
#Allocate data in a dataframe
df <- data.frame(x=x)
#Compute variable
df$y <- sqrt(((df$x^3)+3*((df$x)^2)+1)/((df$x^4)+5*(df$x)^3+7*(df$x)+9))
Finally, we plot:
#Plot
ggplot(df,aes(x=x,y=y))+
geom_point()+
scale_x_continuous(limits = c(0,10))
Output:
Values for x are randomly generated, if you have real values for x you should use those values.