"For the function $f$ it is the case that $f(x) = x^4 – 18x^2$. For which values of $x$ does the graph to the function have a tangent with the slope = 15?"
This is how I did it: Since we're dealing with a tangent of a curve, and its slope, this means that I should take the derivative of this function. E.g. if the slope of the curve on the point $x_t = 5$ then the tangent to that point on the curve has a slope of 5.
$f'(x) = 4x^3 – 36x$
$4x^3 – 36x = 15$
This has no real solutions. The answer in the book is: $x_1 = 0, x_2 = 3, x_3 = -3$.
Where am I going wrong with what I'm doing?
Best Answer
The solutions that your text provided are solutions to the values of $x$ where $f'(x) = 0$, where the line tangent at those values of $x$ is horizontal, not where slope is $15$!
Note:
$$f'(x) = 4x^3 - 36x = x(2x+6)(2x - 6) = 0 \iff x = 0, \;\text{ or}\;x = \pm 3$$
You're derivative is correct, and the equation to solve, as well. The value of slope $=15$ must be a misprint in the text.