Q. 83 - Precalculus Enhanced with Graphing Utilities

Question

If $f\left(x\right)=2x^3+A{x}^2+4x-5$ and $f\left(2\right)=5$, what is the value of $A$?

Step-by-Step Solution

Answer

The value of $A=-\frac{7}{2}$.

Step 1. Given Information

If $f\left(x\right)=2x^3+A{x}^2+4x-5$ and $f\left(2\right)=5$

We have to find the value of $A$.

Step 2. To find the value of A we firstly find the f ( 2 ) by putting x = 2 in the given function.

$$\begin{gathered} f\left(2\right)=2{\left(2\right)}^3+A{\left(2\right)}^2+4·2-5 \\ f\left(2\right)=2·8+A·4+8-5 \\ f\left(2\right)=16+4A+8-5 \\ f\left(2\right)=19+4A \end{gathered}$$

Step 3. Now the function is f ( 2 ) = 19 + 4 A .

We know $f\left(2\right)=5$ so we put the value of $f\left(2\right)$ in the function.

$5=19+4A$

Subtract 19 on both side

$$\begin{gathered} 5-19=19+4A-19 \\ -14=4A \\ 4A=-14 \\ \frac{4}{4}A=-\frac{14}{4} \\ A=-\frac{7}{2} \end{gathered}$$