text: # Find the absolute min and max value on given interval
  
  2,439
  
  The first issue is that $-2 \sqrt{2}$ is not in the interval [-1,4]. However, $2\sqrt{2}$ is indeed a critical point, as $$\frac{\mathrm{d}f}{\mathrm{d}t} = \sqrt{16-t^2} -\frac{t^2}{\sqrt{16-t^2}} = \frac{16-2t^2}{\sqrt{16-t^2}}$$ which is zero at $t=2\sqrt{2}$. Now our possible candidates for the absolute minima and maxima are $t = -1, 2\sqrt{2}, 4$. Evaulating, we obtain $f(-1) = -\sqrt{16-1} = -\sqrt{15}$, $f(2\sqrt{2}) = 2\sqrt{2}(\sqrt{16-8}) = 8$ and $f(4) = 4\sqrt{16-16} = 0$. Thus, our absolute minimum value is $-\sqrt{15}$ at $t=-1$ and our absolute maximum value is $8$ at $t=2\sqrt{2}$.
  
  Share:
  2,439
  
  Author by
  
  ### user106039
  
  Updated on January 29, 2020
  
  $f(t) = t\sqrt{16 − t^2}$ on interval $[−1, 4]$. I was able to find the critical points which are 2√2 and -2√2. All was left for me to do was to evaluate both my critical and end points into the given function. I ended up with the points: $(-1,-3.87) ; (4,0) ; (2\sqrt2,33.94)$ and $(-2\sqrt2,-8)$
  I thought the absolute minimum value had to be $(-2\sqrt2,-8)$ and the absolute maximum value was $(2\sqrt2,33.94)$ but apparently this is incorrect. Where did I go wrong?
metadata: {"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": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}