0
$\begingroup$

Let $$f(x)= 2x^3 +Ax^2 +4x -5$$ Find $A$ given $f(2) =5$.

If I could have someone show me how to solve this it would be great

$\endgroup$
5
$\begingroup$

Hint: The solution is very straightforward. Since we know that $f(2) = 5$, then obviously, $$f(2) = 2(2)^3 + A(2)^2 + 4(2) - 5 = 5$$ $$\implies 16 + 4A + 8 - 5 = 5$$ $$A = \cdots?$$

$\endgroup$
  • 1
    $\begingroup$ Ohhhh so A=-7/2 $\endgroup$ – Ricardo Jan 31 at 0:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.