# Thread:

1. ## Tangent Line

Find the equation of the line tangent to the graph of $f(x) = -4x^3 + 3\sqrt[5]{x} +4$ at the point $(-1,f(-1))$. Leave your answer in the form $y = mx + b$.

I just want to verify if what I'm doing is correct... It has been awhile since I have done one of these.

I assumed I should use the equation $y = f(a) + f'(a)(x-a)$

Which gave me:

$f(-1) = 8 + 3\sqrt[5]{-1}$
$f'(-1) = 12 + \frac{3}{5}\sqrt[5]{(-1)^4} = \frac{63}{5}$

$y = 8 + 3\sqrt[5]{-1} + \frac{63}{5}(x+1)$
$y = \frac{63}{5}x + (8 + \frac{63}{5} + 3\sqrt[5]{-1})$
$y = \frac{63}{5}x + (\frac{103}{5} + 3\sqrt[5]{-1})$

Here is where I have trouble... I am unsure of the fifth root of -1. I have seen that it could be complex, or it could be 1... And I just can't verify it myself. I'm sure it is easier than I am making it out to be, but any help would be appreciated.

2. ## Re: Tangent Line

In the real numbers $\sqrt[5]{-1}=-1$.
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9672760367393494, "perplexity_flag": "head"}