Q. 2 - Calculus

Question

 Solving function equations: Solve each of the following equations for x, and illustrate these solutions on a graph of y = f(x). 

 If $f\left(x\right)=x^3$, solve f(x) = 7.5 and f(x) = 8.5. 

If $f\left(x\right)=\sqrt{x-1}$, solve f(x) = 1.8 and f(x) = 2.2. 

 If $f\left(x\right)=-0.033058\left(x^2-110x\right)$, solve f(x) = 90. 

 If $f\left(x\right)=2-x^2\text{ and }x>0$, solve f(x) = −7.01 and f(x) = −6.99. 

 If $f\left(x\right)=\frac{x^2-2x-3}{x-1}\text{ and }x<0$, solve f(x) = −2 and f(x) = 0. 

Step-by-Step Solution

Answer

Ans:  

(a)  The value of $x=1.95, 2.04$

(b)  The value of $x=4.24, 5.84$

(c)  The value of $x=37.607, 72.392$

(d)  The value of $x=3.001, 2.998$

(e)  The value of $x=-\sqrt{5}, -1$

Step 1. Given information.

given,

     The solution of the equation represents the solution on the graph.

Step 2. (a) Value of x when f(x)=8.5

     $f\left(x\right)=x^3$

     $x=1.95$

$\begin{array}{r}{x}^3=8.5 \\ x=2.040828\end{array}$

Step 3. (b) The solution of the equation represents the solution on the graph

 $f\left(x\right)=\sqrt{x-1}$

Now the value of x when $f\left(x\right)=1.8$

 $\begin{array}{r}\sqrt{x-1}=1.8\\ x=4.2\end{array}$

Now, the value of x when $f\left(x\right)=2.2$

  $\begin{array}{r}\sqrt{x-1}=2.2\\ x=5.84\end{array}$

Step 4. (c) The solution of the equation represents the solution on the graph

   $f\left(x\right)=-0.033058\left(x^2-110x\right)$

Now value of x when f(x)=90

  $\begin{array}{r}-0.033058\left(x^2-110x\right)=90\\ x=37.6071,72.392\end{array}$

 

Step 5. (d) The solution of the equation represents the solution on the graph

 $f\left(x\right)=2-x^2$

Now, the value of x when $f\left(x\right)=-7.01$

 $\begin{array}{r}2-x^2=-7.01\\ x=3.001,-3.001\end{array}$

It is given that $x>0$

   $x=3.001$

Value of x when $f\left(x\right)=-6.99$

  $\begin{array}{r}2-x^2=-6.99\\ x=2.99833,-2.998333\end{array}$

It is given that $x>0$

  $x=2.99833$

Step 6. The solution of the equation represents the solution on the graph

 $f\left(x\right)=\frac{x^2-2x-3}{x-1}$

Now, the value of x when $f\left(x\right)=-2$

  $\begin{array}{r}\frac{x^2-2x-3}{x-1}=-2\\ x=\sqrt{5},-\sqrt{5}\end{array}$

It is given that $x<0$

  $x=-\sqrt{5}$

Value of x when f(x)=0

   $\begin{array}{r}\frac{x^2-2x-3}{x-1}=0\\ x=-1,3\end{array}$

It is given that $x<0$

   $x=-1$